TYPICAL POINTERS WILL BE SIMPLE FOR YOU FROM NOW


What is a pointer?
One of those things beginners in C find difficult is the concept of pointers. The purpose

of this
tutorial is to provide an introduction to pointers and their use to these beginners.
I have found that often the main reason beginners have a problem with pointers is that

they have a
weak or minimal feeling for variables, (as they are used in C). Thus we start with a

discussion of C
variables in general.
A variable in a program is something with a name, the value of which can vary. The way

the
compiler and linker handles this is that it assigns a specific block of memory within the

computer to
hold the value of that variable. The size of that block depends on the range over which

the variable
is allowed to vary. For example, on 32 bit PC's the size of an integer variable is 4

bytes. On older 16
bit PCs integers were 2 bytes. In C the size of a variable type such as an integer need

not be the
same on all types of machines. Further more there is more than one type of integer

variable in C. We
have integers, long integers and short integers which you can read up on in any basic

text on C. This
document assumes the use of a 32 bit system with 4 byte integers.
If you want to know the size of the various types of integers on your system, running the

following
code will give you that information.
#include <stdio.h>
int main()
{
printf("size of a short is %d
", sizeof(short));
printf("size of a int is %d
", sizeof(int));
printf("size of a long is %d
", sizeof(long));
}
When we declare a variable we inform the compiler of two things, the name of the variable

and the
type of the variable. For example, we declare a variable of type integer with the name k

by writing:
int k;
On seeing the "int" part of this statement the compiler sets aside 4 bytes of memory (on

a PC) to
hold the value of the integer. It also sets up a symbol table. In that table it adds the

symbol k and the
relative address in memory where those 4 bytes were set aside.
Thus, later if we write:
k = 2;
we expect that, at run time when this statement is executed, the value 2 will be placed

in that
memory location reserved for the storage of the value of k. In C we refer to a variable

such as the
integer k as an "object".
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In a sense there are two "values" associated with the object k. One is the value of the

integer stored
there (2 in the above example) and the other the "value" of the memory location, i.e.,

the address of
k. Some texts refer to these two values with the nomenclature rvalue (right value,

pronounced "are
value") and lvalue (left value, pronounced "el value") respectively.
In some languages, the lvalue is the value permitted on the left side of the assignment

operator
'=' (i.e. the address where the result of evaluation of the right side ends up). The

rvalue is that which
is on the right side of the assignment statement, the 2 above. Rvalues cannot be used on

the left side
of the assignment statement. Thus: 2 = k; is illegal.
Actually, the above definition of "lvalue" is somewhat modified for C. According to K&R

II (page
197): [1]
"An object is a named region of storage; an lvalue is an expression referring to an
object."
However, at this point, the definition originally cited above is sufficient. As we become

more
familiar with pointers we will go into more detail on this.
Okay, now consider:
int j, k;
k = 2;
j = 7; <-- line 1
k = j; <-- line 2
In the above, the compiler interprets the j in line 1 as the address of the variable j

(its lvalue) and
creates code to copy the value 7 to that address. In line 2, however, the j is

interpreted as its rvalue
(since it is on the right hand side of the assignment operator '='). That is, here the j

refers to the
value stored at the memory location set aside for j, in this case 7. So, the 7 is copied

to the address
designated by the lvalue of k.
In all of these examples, we are using 4 byte integers so all copying of rvalues from one

storage
location to the other is done by copying 4 bytes. Had we been using two byte integers, we

would be
copying 2 bytes.
Now, let's say that we have a reason for wanting a variable designed to hold an lvalue

(an address).
The size required to hold such a value depends on the system. On older desk top computers

with
64K of memory total, the address of any point in memory can be contained in 2 bytes.

Computers
with more memory would require more bytes to hold an address. The actual size required is

not too
important so long as we have a way of informing the compiler that what we want to store

is an
address.
Such a variable is called a pointer variable (for reasons which hopefully will become

clearer a little
later). In C when we define a pointer variable we do so by preceding its name with an

asterisk. In C
we also give our pointer a type which, in this case, refers to the type of data stored at

the address we
will be storing in our pointer. For example, consider the variable declaration:
int *ptr;
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ptr is the name of our variable (just as k was the name of our integer variable). The '*'

informs the
compiler that we want a pointer variable, i.e. to set aside however many bytes is

required to store an
address in memory. The int says that we intend to use our pointer variable to store the

address of an
integer. Such a pointer is said to "point to" an integer. However, note that when we

wrote int k; we
did not give k a value. If this definition is made outside of any function ANSI compliant

compilers
will initialize it to zero. Similarly, ptr has no value, that is we haven't stored an

address in it in the
above declaration. In this case, again if the declaration is outside of any function, it

is initialized to a
value guaranteed in such a way that it is guaranteed to not point to any C object or

function. A
pointer initialized in this manner is called a "null" pointer.
The actual bit pattern used for a null pointer may or may not evaluate to zero since it

depends on the
specific system on which the code is developed. To make the source code compatible

between
various compilers on various systems, a macro is used to represent a null pointer. That

macro goes
under the name NULL. Thus, setting the value of a pointer using the NULL macro, as with

an
assignment statement such as ptr = NULL, guarantees that the pointer has become a null

pointer.
Similarly, just as one can test for an integer value of zero, as in if(k == 0), we can

test for a null
pointer using if (ptr == NULL).
But, back to using our new variable ptr. Suppose now that we want to store in ptr the

address of our
integer variable k. To do this we use the unary & operator and write:
ptr = &k;
What the & operator does is retrieve the lvalue (address) of k, even though k is on the

right hand
side of the assignment operator '=', and copies that to the contents of our pointer ptr.

Now, ptr is said
to "point to" k. Bear with us now, there is only one more operator we need to discuss.
The "dereferencing operator" is the asterisk and it is used as follows:
*ptr = 7;
will copy 7 to the address pointed to by ptr. Thus if ptr "points to" (contains the

address of) k, the
above statement will set the value of k to 7. That is, when we use the '*' this way we

are referring to
the value of that which ptr is pointing to, not the value of the pointer itself.
Similarly, we could write:
printf("%d
",*ptr);
to print to the screen the integer value stored at the address pointed to by ptr;.
One way to see how all this stuff fits together would be to run the following program and

then
review the code and the output carefully.
------------ Program 1.1 ---------------------------------
/* Program 1.1 from PTRTUT10.TXT 6/10/97 */
#include
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int j, k;
int *ptr;
int main(void)
{
j = 1;
k = 2;
ptr = &k;
printf("
");
printf("j has the value %d and is stored at %p
", j, (void *)&j);
printf("k has the value %d and is stored at %p
", k, (void *)&k);
printf("ptr has the value %p and is stored at %p
", ptr, (void *)&ptr);
printf("The value of the integer pointed to by ptr is %d
", *ptr);
return 0;
}
Note: We have yet to discuss those aspects of C which require the use of the (void *)

expression
used here. For now, include it in your test code. We'll explain the reason behind this

expression
later.
To review:
A variable is declared by giving it a type and a name (e.g. int k;)
A pointer variable is declared by giving it a type and a name (e.g. int *ptr) where the

asterisk
tells the compiler that the variable named ptr is a pointer variable and the type tells

the
compiler what type the pointer is to point to (integer in this case).
Once a variable is declared, we can get its address by preceding its name with the

unary &
operator, as in &k.
We can "dereference" a pointer, i.e. refer to the value of that which it points to, by

using the
unary '*' operator as in *ptr.
An "lvalue" of a variable is the value of its address, i.e. where it is stored in

memory. The
"rvalue" of a variable is the value stored in that variable (at that address).
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Pointer types and Arrays
Okay, let's move on. Let us consider why we need to identify the type of variable that a

pointer
points to, as in:
int *ptr;
One reason for doing this is so that later, once ptr "points to" something, if we write:
*ptr = 2;
the compiler will know how many bytes to copy into that memory location pointed to by

ptr. If ptr
was declared as pointing to an integer, 4 bytes would be copied. Similarly for floats and

doubles the
appropriate number will be copied. But, defining the type that the pointer points to

permits a number
of other interesting ways a compiler can interpret code. For example, consider a block in

memory
consisting of ten integers in a row. That is, 40 bytes of memory are set aside to hold 10

integers.
Now, let's say we point our integer pointer ptr at the first of these integers.

Furthermore lets say that
integer is located at memory location 100 (decimal). What happens when we write:
ptr + 1;
Because the compiler "knows" this is a pointer (i.e. its value is an address) and that it

points to an
integer (its current address, 100, is the address of an integer), it adds 4 to ptr

instead of 1, so the
pointer "points to" the next integer, at memory location 104. Similarly, were the ptr

declared as a
pointer to a short, it would add 2 to it instead of 1. The same goes for other data types

such as floats,
doubles, or even user defined data types such as structures. This is obviously not the

same kind of
"addition" that we normally think of. In C it is referred to as addition using "pointer

arithmetic", a
term which we will come back to later.
Similarly, since ++ptr and ptr++ are both equivalent to ptr + 1 (though the point in the

program
when ptr is incremented may be different), incrementing a pointer using the unary ++

operator,
either pre- or post-, increments the address it stores by the amount sizeof(type) where

"type" is the
type of the object pointed to. (i.e. 4 for an integer).
Since a block of 10 integers located contiguously in memory is, by definition, an array

of integers,
this brings up an interesting relationship between arrays and pointers.
Consider the following:
int my_array[] = {1,23,17,4,-5,100};
Here we have an array containing 6 integers. We refer to each of these integers by means

of a
subscript to my_array, i.e. using my_array[0] through my_array[5]. But, we could

alternatively
access them via a pointer as follows:
int *ptr;
ptr = &my_array[0]; /* point our pointer at the first
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integer in our array */
And then we could print out our array either using the array notation or by dereferencing

our
pointer. The following code illustrates this:
----------- Program 2.1 -----------------------------------
/* Program 2.1 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
int my_array[] = {1,23,17,4,-5,100};
int *ptr;
int main(void)
{
int i;
ptr = &my_array[0]; /* point our pointer to the first
element of the array */
printf("
");
for (i = 0; i < 6; i++)
{
printf("my_array[%d] = %d ",i,my_array[i]); /*<-- A */
printf("ptr + %d = %d
",i, *(ptr + i)); /*<-- B */
}
return 0;
}
Compile and run the above program and carefully note lines A and B and that the program

prints out
the same values in either case. Also observe how we dereferenced our pointer in line B,

i.e. we first
added i to it and then dereferenced the new pointer. Change line B to read:
printf("ptr + %d = %d
",i, *ptr++);
and run it again... then change it to:
printf("ptr + %d = %d
",i, *(++ptr));
and try once more. Each time try and predict the outcome and carefully look at the actual

outcome.
In C, the standard states that wherever we might use &var_name[0] we can replace that

with
var_name, thus in our code where we wrote:
ptr = &my_array[0];
we can write:
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ptr = my_array;
to achieve the same result.
This leads many texts to state that the name of an array is a pointer. I prefer to

mentally think "the
name of the array is the address of first element in the array". Many beginners

(including myself
when I was learning) have a tendency to become confused by thinking of it as a pointer.

For
example, while we can write
ptr = my_array;
we cannot write
my_array = ptr;
The reason is that while ptr is a variable, my_array is a constant. That is, the location

at which the
first element of my_array will be stored cannot be changed once my_array[] has been

declared.
Earlier when discussing the term "lvalue" I cited K&R-2 where it stated:
"An object is a named region of storage; an lvalue is an expression referring to an
object".
This raises an interesting problem. Since my_array is a named region of storage, why is

my_array
in the above assignment statement not an lvalue? To resolve this problem, some refer to

my_array
as an "unmodifiable lvalue".
Modify the example program above by changing
ptr = &my_array[0];
to
ptr = my_array;
and run it again to verify the results are identical.
Now, let's delve a little further into the difference between the names ptr and my_array

as used
above. Some writers will refer to an array's name as a constant pointer. What do we mean

by that?
Well, to understand the term "constant" in this sense, let's go back to our definition of

the term
"variable". When we declare a variable we set aside a spot in memory to hold the value of

the
appropriate type. Once that is done the name of the variable can be interpreted in one of

two ways.
When used on the left side of the assignment operator, the compiler interprets it as the

memory
location to which to move that value resulting from evaluation of the right side of the

assignment
operator. But, when used on the right side of the assignment operator, the name of a

variable is
interpreted to mean the contents stored at that memory address set aside to hold the

value of that
variable.
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With that in mind, let's now consider the simplest of constants, as in:
int i, k;
i = 2;
Here, while i is a variable and then occupies space in the data portion of memory, 2 is a

constant
and, as such, instead of setting aside memory in the data segment, it is imbedded

directly in the code
segment of memory. That is, while writing something like k = i; tells the compiler to

create code
which at run time will look at memory location &i to determine the value to be moved to

k, code
created by i = 2; simply puts the 2 in the code and there is no referencing of the data

segment. That
is, both k and i are objects, but 2 is not an object.
Similarly, in the above, since my_array is a constant, once the compiler establishes

where the array
itself is to be stored, it "knows" the address of my_array[0] and on seeing:
ptr = my_array;
it simply uses this address as a constant in the code segment and there is no referencing

of the data
segment beyond that.
This might be a good place explain further the use of the (void *) expression used in

Program 1.1 of
Chapter 1. As we have seen we can have pointers of various types. So far we have

discussed
pointers to integers and pointers to characters. In coming chapters we will be learning

about pointers
to structures and even pointer to pointers.
Also we have learned that on different systems the size of a pointer can vary. As it

turns out it is also
possible that the size of a pointer can vary depending on the data type of the object to

which it
points. Thus, as with integers where you can run into trouble attempting to assign a long

integer to a
variable of type short integer, you can run into trouble attempting to assign the values

of pointers of
various types to pointer variables of other types.
To minimize this problem, C provides for a pointer of type void. We can declare such a

pointer by
writing:
void *vptr;
A void pointer is sort of a generic pointer. For example, while C will not permit the

comparison of a
pointer to type integer with a pointer to type character, for example, either of these

can be compared
to a void pointer. Of course, as with other variables, casts can be used to convert from

one type of
pointer to another under the proper circumstances. In Program 1.1. of Chapter 1 I cast

the pointers to
integers into void pointers to make them compatible with the %p conversion specification.

In later
chapters other casts will be made for reasons defined therein.
Well, that's a lot of technical stuff to digest and I don't expect a beginner to

understand all of it on
first reading. With time and experimentation you will want to come back and re-read the

first 2
chapters. But for now, let's move on to the relationship between pointers, character

arrays, and
strings.
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Pointers and Strings
The study of strings is useful to further tie in the relationship between pointers and

arrays. It also
makes it easy to illustrate how some of the standard C string functions can be

implemented. Finally
it illustrates how and when pointers can and should be passed to functions.
In C, strings are arrays of characters. This is not necessarily true in other languages.

In BASIC,
Pascal, Fortran and various other languages, a string has its own data type. But in C it

does not. In C
a string is an array of characters terminated with a binary zero character (written as

''). To start off
our discussion we will write some code which, while preferred for illustrative purposes,

you would
probably never write in an actual program. Consider, for example:
char my_string[40];
my_string[0] = 'T';
my_string[1] = 'e';
my_string[2] = 'd':
my_string[3] = '';
While one would never build a string like this, the end result is a string in that it is

an array of
characters terminated with a nul character. By definition, in C, a string is an array of

characters
terminated with the nul character. Be aware that "nul" is not the same as "NULL". The nul

refers to
a zero as defined by the escape sequence ''. That is it occupies one byte of memory.

NULL, on the
other hand, is the name of the macro used to initialize null pointers. NULL is #defined

in a header
file in your C compiler, nul may not be #defined at all.
Since writing the above code would be very time consuming, C permits two alternate ways

of
achieving the same thing. First, one might write:
char my_string[40] = {'T', 'e', 'd', '',};
But this also takes more typing than is convenient. So, C permits:
char my_string[40] = "Ted";
When the double quotes are used, instead of the single quotes as was done in the previous

examples,
the nul character ( '' ) is automatically appended to the end of the string.
In all of the above cases, the same thing happens. The compiler sets aside an contiguous

block of
memory 40 bytes long to hold characters and initialized it such that the first 4

characters are Ted.
Now, consider the following program:
------------------program 3.1-------------------------------------
/* Program 3.1 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
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char strA[80] = "A string to be used for demonstration purposes";
char strB[80];
int main(void)
{
char *pA; /* a pointer to type character */
char *pB; /* another pointer to type character */
puts(strA); /* show string A */
pA = strA; /* point pA at string A */
puts(pA); /* show what pA is pointing to */
pB = strB; /* point pB at string B */
putchar('
'); /* move down one line on the screen */
while(*pA != '') /* line A (see text) */
{
*pB++ = *pA++; /* line B (see text) */
}
*pB = ''; /* line C (see text) */
puts(strB); /* show strB on screen */
return 0;
}
--------- end program 3.1 -------------------------------------
In the above we start out by defining two character arrays of 80 characters each. Since

these are
globally defined, they are initialized to all ''s first. Then, strA has the first 42

characters initialized
to the string in quotes.
Now, moving into the code, we declare two character pointers and show the string on the

screen. We
then "point" the pointer pA at strA. That is, by means of the assignment statement we

copy the
address of strA[0] into our variable pA. We now use puts() to show that which is pointed

to by pA
on the screen. Consider here that the function prototype for puts() is:
int puts(const char *s);
For the moment, ignore the const. The parameter passed to puts() is a pointer, that is

the value of a
pointer (since all parameters in C are passed by value), and the value of a pointer is

the address to
which it points, or, simply, an address. Thus when we write puts(strA); as we have seen,

we are
passing the address of strA[0].
Similarly, when we write puts(pA); we are passing the same address, since we have set pA

= strA;
Given that, follow the code down to the while() statement on line A. Line A states:
While the character pointed to by pA (i.e. *pA) is not a nul character (i.e. the

terminating ''), do the
following:
Line B states: copy the character pointed to by pA to the space pointed to by pB, then

increment pA
so it points to the next character and pB so it points to the next space.
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When we have copied the last character, pA now points to the terminating nul character

and the loop
ends. However, we have not copied the nul character. And, by definition a string in C

must be nul
terminated. So, we add the nul character with line C.
It is very educational to run this program with your debugger while watching strA, strB,

pA and pB
and single stepping through the program. It is even more educational if instead of simply

defining
strB[] as has been done above, initialize it also with something like:
strB[80] = "12345678901234567890123456789012345678901234567890"
where the number of digits used is greater than the length of strA and then repeat the

single stepping
procedure while watching the above variables. Give these things a try!
Getting back to the prototype for puts() for a moment, the "const" used as a parameter

modifier
informs the user that the function will not modify the string pointed to by s, i.e. it

will treat that
string as a constant.
Of course, what the above program illustrates is a simple way of copying a string. After

playing with
the above until you have a good understanding of what is happening, we can proceed to

creating our
own replacement for the standard strcpy() that comes with C. It might look like:
char *my_strcpy(char *destination, char *source)
{
char *p = destination;
while (*source != '')
{
*p++ = *source++;
}
*p = '';
return destination;
}
In this case, I have followed the practice used in the standard routine of returning a

pointer to the
destination.
Again, the function is designed to accept the values of two character pointers, i.e.

addresses, and
thus in the previous program we could write:
int main(void)
{
my_strcpy(strB, strA);
puts(strB);
}
I have deviated slightly from the form used in standard C which would have the prototype:
char *my_strcpy(char *destination, const char *source);
Here the "const" modifier is used to assure the user that the function will not modify

the contents
pointed to by the source pointer. You can prove this by modifying the function above, and

its
prototype, to include the "const" modifier as shown. Then, within the function you can

add a
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statement which attempts to change the contents of that which is pointed to by source,

such as:
*source = 'X';
which would normally change the first character of the string to an X. The const modifier

should
cause your compiler to catch this as an error. Try it and see.
Now, let's consider some of the things the above examples have shown us. First off,

consider the
fact that *ptr++ is to be interpreted as returning the value pointed to by ptr and then

incrementing
the pointer value. This has to do with the precedence of the operators. Were we to write

(*ptr)++ we
would increment, not the pointer, but that which the pointer points to! i.e. if used on

the first
character of the above example string the 'T' would be incremented to a 'U'. You can

write some
simple example code to illustrate this.
Recall again that a string is nothing more than an array of characters, with the last

character being a
''. What we have done above is deal with copying an array. It happens to be an array of

characters
but the technique could be applied to an array of integers, doubles, etc. In those cases,

however, we
would not be dealing with strings and hence the end of the array would not be marked with

a special
value like the nul character. We could implement a version that relied on a special value

to identify
the end. For example, we could copy an array of positive integers by marking the end with

a
negative integer. On the other hand, it is more usual that when we write a function to

copy an array
of items other than strings we pass the function the number of items to be copied as well

as the
address of the array, e.g. something like the following prototype might indicate:
void int_copy(int *ptrA, int *ptrB, int nbr);
where nbr is the number of integers to be copied. You might want to play with this idea

and create
an array of integers and see if you can write the function int_copy() and make it work.
This permits using functions to manipulate large arrays. For example, if we have an array

of 5000
integers that we want to manipulate with a function, we need only pass to that function

the address
of the array (and any auxiliary information such as nbr above, depending on what we are

doing).
The array itself does not get passed, i.e. the whole array is not copied and put on the

stack before
calling the function, only its address is sent.
This is different from passing, say an integer, to a function. When we pass an integer we

make a
copy of the integer, i.e. get its value and put it on the stack. Within the function any

manipulation of
the value passed can in no way effect the original integer. But, with arrays and pointers

we can pass
the address of the variable and hence manipulate the values of the original variables.
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More on Strings
Well, we have progressed quite a way in a short time! Let's back up a little and look at

what was
done in Chapter 3 on copying of strings but in a different light. Consider the following

function:
char *my_strcpy(char dest[], char source[])
{
int i = 0;
while (source[i] != '')
{
dest[i] = source[i];
i++;
}
dest[i] = '';
return dest;
}
Recall that strings are arrays of characters. Here we have chosen to use array notation

instead of
pointer notation to do the actual copying. The results are the same, i.e. the string gets

copied using
this notation just as accurately as it did before. This raises some interesting points

which we will
discuss.
Since parameters are passed by value, in both the passing of a character pointer or the

name of the
array as above, what actually gets passed is the address of the first element of each

array. Thus, the
numerical value of the parameter passed is the same whether we use a character pointer or

an array
name as a parameter. This would tend to imply that somehow source[i] is the same as *

(p+i).
In fact, this is true, i.e wherever one writes a[i] it can be replaced with *(a + i)

without any
problems. In fact, the compiler will create the same code in either case. Thus we see

that pointer
arithmetic is the same thing as array indexing. Either syntax produces the same result.
This is NOT saying that pointers and arrays are the same thing, they are not. We are only

saying that
to identify a given element of an array we have the choice of two syntaxes, one using

array indexing
and the other using pointer arithmetic, which yield identical results.
Now, looking at this last expression, part of it.. (a + i), is a simple addition using

the + operator and
the rules of C state that such an expression is commutative. That is (a + i) is identical

to (i + a). Thus
we could write *(i + a) just as easily as *(a + i).
But *(i + a) could have come from i[a] ! From all of this comes the curious truth that

if:
char a[20];
int i;
writing
a[3] = 'x';
is the same as writing
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3[a] = 'x';
Try it! Set up an array of characters, integers or longs, etc. and assigned the 3rd or

4th element a
value using the conventional approach and then print out that value to be sure you have

that
working. Then reverse the array notation as I have done above. A good compiler will not

balk and
the results will be identical. A curiosity... nothing more!
Now, looking at our function above, when we write:
dest[i] = source[i];
due to the fact that array indexing and pointer arithmetic yield identical results, we

can write this as:
*(dest + i) = *(source + i);
But, this takes 2 additions for each value taken on by i. Additions, generally speaking,

take more
time than incrementations (such as those done using the ++ operator as in i++). This may

not be true
in modern optimizing compilers, but one can never be sure. Thus, the pointer version may

be a bit
faster than the array version.
Another way to speed up the pointer version would be to change:
while (*source != '')
to simply
while (*source)
since the value within the parenthesis will go to zero (FALSE) at the same time in either

case.
At this point you might want to experiment a bit with writing some of your own programs

using
pointers. Manipulating strings is a good place to experiment. You might want to write

your own
versions of such standard functions as:
strlen();
strcat();
strchr();
and any others you might have on your system.
We will come back to strings and their manipulation through pointers in a future chapter.

For now,
let's move on and discuss structures for a bit.
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Pointers and Structures
As you may know, we can declare the form of a block of data containing different data

types by
means of a structure declaration. For example, a personnel file might contain structures

which look
something like:
struct tag {
char lname[20]; /* last name */
char fname[20]; /* first name */
int age; /* age */
float rate; /* e.g. 12.75 per hour */
};
Let's say we have a bunch of these structures in a disk file and we want to read each one

out and
print out the first and last name of each one so that we can have a list of the people in

our files. The
remaining information will not be printed out. We will want to do this printing with a

function call
and pass to that function a pointer to the structure at hand. For demonstration purposes

I will use
only one structure for now. But realize the goal is the writing of the function, not the

reading of the
file which, presumably, we know how to do.
For review, recall that we can access structure members with the dot operator as in:
--------------- program 5.1 ------------------
/* Program 5.1 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <string.h>
struct tag {
char lname[20]; /* last name */
char fname[20]; /* first name */
int age; /* age */
float rate; /* e.g. 12.75 per hour */
};
struct tag my_struct; /* declare the structure my_struct */
int main(void)
{
strcpy(my_struct.lname,"Jensen");
strcpy(my_struct.fname,"Ted");
printf("
%s ",my_struct.fname);
printf("%s
",my_struct.lname);
return 0;
}
-------------- end of program 5.1 --------------
Now, this particular structure is rather small compared to many used in C programs. To

the above
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we might want to add:
date_of_hire; (data types not shown)
date_of_last_raise;
last_percent_increase;
emergency_phone;
medical_plan;
Social_S_Nbr;
etc.....
If we have a large number of employees, what we want to do is manipulate the data in

these
structures by means of functions. For example we might want a function print out the name

of the
employee listed in any structure passed to it. However, in the original C (Kernighan &

Ritchie, 1st
Edition) it was not possible to pass a structure, only a pointer to a structure could be

passed. In
ANSI C, it is now permissible to pass the complete structure. But, since our goal here is

to learn
more about pointers, we won't pursue that.
Anyway, if we pass the whole structure it means that we must copy the contents of the

structure
from the calling function to the called function. In systems using stacks, this is done

by pushing the
contents of the structure on the stack. With large structures this could prove to be a

problem.
However, passing a pointer uses a minimum amount of stack space.
In any case, since this is a discussion of pointers, we will discuss how we go about

passing a pointer
to a structure and then using it within the function.
Consider the case described, i.e. we want a function that will accept as a parameter a

pointer to a
structure and from within that function we want to access members of the structure. For

example we
want to print out the name of the employee in our example structure.
Okay, so we know that our pointer is going to point to a structure declared using struct

tag. We
declare such a pointer with the declaration:
struct tag *st_ptr;
and we point it to our example structure with:
st_ptr = &my_struct;
Now, we can access a given member by de-referencing the pointer. But, how do we de-

reference the
pointer to a structure? Well, consider the fact that we might want to use the pointer to

set the age of
the employee. We would write:
(*st_ptr).age = 63;
Look at this carefully. It says, replace that within the parenthesis with that which

st_ptr points to,
which is the structure my_struct. Thus, this breaks down to the same as my_struct.age.
However, this is a fairly often used expression and the designers of C have created an

alternate
syntax with the same meaning which is:
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st_ptr->age = 63;
With that in mind, look at the following program:
------------ program 5.2 ---------------------
/* Program 5.2 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <string.h>
struct tag{ /* the structure type */
char lname[20]; /* last name */
char fname[20]; /* first name */
int age; /* age */
float rate; /* e.g. 12.75 per hour */
};
struct tag my_struct; /* define the structure */
void show_name(struct tag *p); /* function prototype */
int main(void)
{
struct tag *st_ptr; /* a pointer to a structure */
st_ptr = &my_struct; /* point the pointer to my_struct */
strcpy(my_struct.lname,"Jensen");
strcpy(my_struct.fname,"Ted");
printf("
%s ",my_struct.fname);
printf("%s
",my_struct.lname);
my_struct.age = 63;
show_name(st_ptr); /* pass the pointer */
return 0;
}
void show_name(struct tag *p)
{
printf("
%s ", p->fname); /* p points to a structure */
printf("%s ", p->lname);
printf("%d
", p->age);
}
-------------------- end of program 5.2 ----------------
Again, this is a lot of information to absorb at one time. The reader should compile and

run the
various code snippets and using a debugger monitor things like my_struct and p while

single
stepping through the main and following the code down into the function to see what is

happening.
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Some more on Strings, and Arrays of Strings
Well, let's go back to strings for a bit. In the following all assignments are to be

understood as being
global, i.e. made outside of any function, including main().
We pointed out in an earlier chapter that we could write:
char my_string[40] = "Ted";
which would allocate space for a 40 byte array and put the string in the first 4 bytes

(three for the
characters in the quotes and a 4th to handle the terminating '').
Actually, if all we wanted to do was store the name "Ted" we could write:
char my_name[] = "Ted";
and the compiler would count the characters, leave room for the nul character and store

the total of
the four characters in memory the location of which would be returned by the array name,

in this
case my_name.
In some code, instead of the above, you might see:
char *my_name = "Ted";
which is an alternate approach. Is there a difference between these? The answer is.. yes.

Using the
array notation 4 bytes of storage in the static memory block are taken up, one for each

character and
one for the terminating nul character. But, in the pointer notation the same 4 bytes

required, plus N
bytes to store the pointer variable my_name (where N depends on the system but is usually

a
minimum of 2 bytes and can be 4 or more).
In the array notation, my_name is short for &myname[0] which is the address of the first

element
of the array. Since the location of the array is fixed during run time, this is a

constant (not a
variable). In the pointer notation my_name is a variable. As to which is the better

method, that
depends on what you are going to do within the rest of the program.
Let's now go one step further and consider what happens if each of these declarations are

done
within a function as opposed to globally outside the bounds of any function.
void my_function_A(char *ptr)
{
char a[] = "ABCDE"
.
.
}
void my_function_B(char *ptr)
{
char *cp = "FGHIJ"
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.
.
}
In the case of my_function_A, the content, or value(s), of the array a[] is considered to

be the data.
The array is said to be initialized to the values ABCDE. In the case of my_function_B,

the value of
the pointer cp is considered to be the data. The pointer has been initialized to point to

the string
FGHIJ. In both my_function_A and my_function_B the definitions are local variables and

thus
the string ABCDE is stored on the stack, as is the value of the pointer cp. The string

FGHIJ can be
stored anywhere. On my system it gets stored in the data segment.
By the way, array initialization of automatic variables as I have done in my_function_A

was illegal
in the older K&R C and only "came of age" in the newer ANSI C. A fact that may be

important
when one is considering portability and backwards compatibility.
As long as we are discussing the relationship/differences between pointers and arrays,

let's move on
to multi-dimensional arrays. Consider, for example the array:
char multi[5][10];
Just what does this mean? Well, let's consider it in the following light.
char multi[5][10];
Let's take the underlined part to be the "name" of an array. Then prepending the char and

appending
the [10] we have an array of 10 characters. But, the name multi[5] is itself an array

indicating that
there are 5 elements each being an array of 10 characters. Hence we have an array of 5

arrays of 10
characters each..
Assume we have filled this two dimensional array with data of some kind. In memory, it

might look
as if it had been formed by initializing 5 separate arrays using something like:
multi[0] = {'0','1','2','3','4','5','6','7','8','9'}
multi[1] = {'a','b','c','d','e','f','g','h','i','j'}
multi[2] = {'A','B','C','D','E','F','G','H','I','J'}
multi[3] = {'9','8','7','6','5','4','3','2','1','0'}
multi[4] = {'J','I','H','G','F','E','D','C','B','A'}
At the same time, individual elements might be addressable using syntax such as:
multi[0][3] = '3'
multi[1][7] = 'h'
multi[4][0] = 'J'
Since arrays are contiguous in memory, our actual memory block for the above should look

like:
0123456789abcdefghijABCDEFGHIJ9876543210JIHGFEDCBA
^
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|_____ starting at the address &multi[0][0]
Note that I did not write multi[0] = "0123456789". Had I done so a terminating '' would

have been
implied since whenever double quotes are used a '' character is appended to the

characters contained
within those quotes. Had that been the case I would have had to set aside room for 11

characters per
row instead of 10.
My goal in the above is to illustrate how memory is laid out for 2 dimensional arrays.

That is, this is
a 2 dimensional array of characters, NOT an array of "strings".
Now, the compiler knows how many columns are present in the array so it can interpret

multi + 1 as
the address of the 'a' in the 2nd row above. That is, it adds 10, the number of columns,

to get this
location. If we were dealing with integers and an array with the same dimension the

compiler would
add 10*sizeof(int) which, on my machine, would be 20. Thus, the address of the 9 in the

4th row
above would be &multi[3][0] or *(multi + 3) in pointer notation. To get to the content of

the 2nd
element in the 4th row we add 1 to this address and dereference the result as in
*(*(multi + 3) + 1)
With a little thought we can see that:
*(*(multi + row) + col) and
multi[row][col] yield the same results.
The following program illustrates this using integer arrays instead of character arrays.
------------------- program 6.1 ----------------------
/* Program 6.1 from PTRTUT10.HTM 6/13/97*/
#include <stdio.h>
#define ROWS 5
#define COLS 10
int multi[ROWS][COLS];
int main(void)
{
int row, col;
for (row = 0; row < ROWS; row++)
{
for (col = 0; col < COLS; col++)
{
multi[row][col] = row*col;
}
}
for (row = 0; row < ROWS; row++)
{
for (col = 0; col < COLS; col++)
{
printf("
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%d ",multi[row][col]);
printf("%d ",*(*(multi + row) + col));
}
}
return 0;
}
----------------- end of program 6.1 ---------------------
Because of the double de-referencing required in the pointer version, the name of a 2

dimensional
array is often said to be equivalent to a pointer to a pointer. With a three dimensional

array we
would be dealing with an array of arrays of arrays and some might say its name would be

equivalent
to a pointer to a pointer to a pointer. However, here we have initially set aside the

block of memory
for the array by defining it using array notation. Hence, we are dealing with a constant,

not a
variable. That is we are talking about a fixed address not a variable pointer. The

dereferencing
function used above permits us to access any element in the array of arrays without the

need of
changing the value of that address (the address of multi[0][0] as given by the symbol

multi).
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More on Multi-Dimensional Arrays
In the previous chapter we noted that given
#define ROWS 5
#define COLS 10
int multi[ROWS][COLS];
we can access individual elements of the array multi using either:
multi[row][col]
or
*(*(multi + row) + col)
To understand more fully what is going on, let us replace
*(multi + row)
with X as in:
*(X + col)
Now, from this we see that X is like a pointer since the expression is de-referenced and

we know
that col is an integer. The arithmetic being used here is of a special kind called

"pointer arithmetic".
That means that, since we are talking about an integer array, the address pointed to by

(i.e. value of)
X + col + 1 must be greater than the address X + col by and amount equal to sizeof(int).
Since we know the memory layout for 2 dimensional arrays, we can determine that in the

expression
multi + row as used above, multi + row + 1 must increase by value an amount equal to that

needed
to "point to" the next row, which in this case would be an amount equal to COLS * sizeof

(int).
That says that if the expression *(*(multi + row) + col) is to be evaluated correctly at

run time, the
compiler must generate code which takes into consideration the value of COLS, i.e. the

2nd
dimension. Because of the equivalence of the two forms of expression, this is true

whether we are
using the pointer expression as here or the array expression multi[row][col].
Thus, to evaluate either expression, a total of 5 values must be known:
1. The address of the first element of the array, which is returned by the expression

multi, i.e.,
the name of the array.
2. The size of the type of the elements of the array, in this case sizeof(int).
3. The 2nd dimension of the array
4. The specific index value for the first dimension, row in this case.
5. The specific index value for the second dimension, col in this case.
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Given all of that, consider the problem of designing a function to manipulate the element

values of a
previously declared array. For example, one which would set all the elements of the array

multi to
the value 1.
void set_value(int m_array[][COLS])
{
int row, col;
for (row = 0; row < ROWS; row++)
{
for (col = 0; col < COLS; col++)
{
m_array[row][col] = 1;
}
}
}
And to call this function we would then use:
set_value(multi);
Now, within the function we have used the values #defined by ROWS and COLS that set the

limits
on the for loops. But, these #defines are just constants as far as the compiler is

concerned, i.e. there
is nothing to connect them to the array size within the function. row and col are local

variables, of
course. The formal parameter definition permits the compiler to determine the

characteristics
associated with the pointer value that will be passed at run time. We really don?t need

the first
dimension and, as will be seen later, there are occasions where we would prefer not to

define it
within the parameter definition, out of habit or consistency, I have not used it here.

But, the second
dimension must be used as has been shown in the expression for the parameter. The reason

is that
we need this in the evaluation of m_array[row][col] as has been described. While the

parameter
defines the data type (int in this case) and the automatic variables for row and column

are defined
for loops, only one value can be passed using a single parameter. In this case, that is

the value of
multi as noted in the call statement, i.e. the address of the first element, often

referred to as a pointer
to the array. Thus, the only way we have of informing the compiler of the 2nd dimension

is by
explicitly including it in the parameter definition.
In fact, in general all dimensions of higher order than one are needed when dealing with

multidimensional
arrays. That is if we are talking about 3 dimensional arrays, the 2nd and 3rd dimension
must be specified in the parameter definition.
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Pointers to Arrays
Pointers, of course, can be "pointed at" any type of data object, including arrays. While

that was
evident when we discussed program 3.1, it is important to expand on how we do this when

it comes
to multi-dimensional arrays.
To review, in Chapter 2 we stated that given an array of integers we could point an

integer pointer at
that array using:
int *ptr;
ptr = &my_array[0]; /* point our pointer at the first
integer in our array */
As we stated there, the type of the pointer variable must match the type of the first

element of the
array.
In addition, we can use a pointer as a formal parameter of a function which is designed

to
manipulate an array. e.g.
Given:
int array[3] = {'1', '5', '7'};
void a_func(int *p);
Some programmers might prefer to write the function prototype as:
void a_func(int p[]);
which would tend to inform others who might use this function that the function is

designed to
manipulate the elements of an array. Of course, in either case, what actually gets passed

is the value
of a pointer to the first element of the array, independent of which notation is used in

the function
prototype or definition. Note that if the array notation is used, there is no need to

pass the actual
dimension of the array since we are not passing the whole array, only the address to the

first
element.
We now turn to the problem of the 2 dimensional array. As stated in the last chapter, C

interprets a 2
dimensional array as an array of one dimensional arrays. That being the case, the first

element of a 2
dimensional array of integers is a one dimensional array of integers. And a pointer to a

two
dimensional array of integers must be a pointer to that data type. One way of

accomplishing this is
through the use of the keyword "typedef". typedef assigns a new name to a specified data

type. For
example:
typedef unsigned char byte;
causes the name byte to mean type unsigned char. Hence
byte b[10]; would be an array of unsigned characters.
Note that in the typedef declaration, the word byte has replaced that which would

normally be the
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name of our unsigned char. That is, the rule for using typedef is that the new name for

the data
type is the name used in the definition of the data type. Thus in:
typedef int Array[10];
Array becomes a data type for an array of 10 integers. i.e. Array my_arr; declares my_arr

as an
array of 10 integers and Array arr2d[5]; makes arr2d an array of 5 arrays of 10 integers

each.
Also note that Array *p1d; makes p1d a pointer to an array of 10 integers. Because *p1d

points to
the same type as arr2d, assigning the address of the two dimensional array arr2d to p1d,

the
pointer to a one dimensional array of 10 integers is acceptable. i.e. p1d = &arr2d[0]; or

p1d =
arr2d; are both correct.
Since the data type we use for our pointer is an array of 10 integers we would expect

that
incrementing p1d by 1 would change its value by 10*sizeof(int), which it does. That is,

sizeof
(*p1d) is 20. You can prove this to yourself by writing and running a simple short

program.
Now, while using typedef makes things clearer for the reader and easier on the

programmer, it is not
really necessary. What we need is a way of declaring a pointer like p1d without the need

of the
typedef keyword. It turns out that this can be done and that
int (*p1d)[10];
is the proper declaration, i.e. p1d here is a pointer to an array of 10 integers just as

it was under the
declaration using the Array type. Note that this is different from
int *p1d[10];
which would make p1d the name of an array of 10 pointers to type int.
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Pointers and Dynamic Allocation of Memory
There are times when it is convenient to allocate memory at run time using malloc(),

calloc(), or
other allocation functions. Using this approach permits postponing the decision on the

size of the
memory block need to store an array, for example, until run time. Or it permits using a

section of
memory for the storage of an array of integers at one point in time, and then when that

memory is no
longer needed it can be freed up for other uses, such as the storage of an array of

structures.
When memory is allocated, the allocating function (such as malloc(), calloc(), etc.)

returns a
pointer. The type of this pointer depends on whether you are using an older K&R compiler

or the
newer ANSI type compiler. With the older compiler the type of the returned pointer is

char, with
the ANSI compiler it is void.
If you are using an older compiler, and you want to allocate memory for an array of

integers you
will have to cast the char pointer returned to an integer pointer. For example, to

allocate space for 10
integers we might write:
int *iptr;
iptr = (int *)malloc(10 * sizeof(int));
if (iptr == NULL)
{ .. ERROR ROUTINE GOES HERE .. }
If you are using an ANSI compliant compiler, malloc() returns a void pointer and since a

void
pointer can be assigned to a pointer variable of any object type, the (int *) cast shown

above is not
needed. The array dimension can be determined at run time and is not needed at compile

time. That
is, the 10 above could be a variable read in from a data file or keyboard, or calculated

based on some
need, at run time.
Because of the equivalence between array and pointer notation, once iptr has been

assigned as
above, one can use the array notation. For example, one could write:
int k;
for (k = 0; k < 10; k++)
iptr[k] = 2;
to set the values of all elements to 2.
Even with a reasonably good understanding of pointers and arrays, one place the newcomer

to C is
likely to stumble at first is in the dynamic allocation of multi-dimensional arrays. In

general, we
would like to be able to access elements of such arrays using array notation, not pointer

notation,
wherever possible. Depending on the application we may or may not know both dimensions at
compile time. This leads to a variety of ways to go about our task.
As we have seen, when dynamically allocating a one dimensional array its dimension can be
determined at run time. Now, when using dynamic allocation of higher order arrays, we

never need
to know the first dimension at compile time. Whether we need to know the higher

dimensions
depends on how we go about writing the code. Here I will discuss various methods of

dynamically
allocating room for 2 dimensional arrays of integers.
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First we will consider cases where the 2nd dimension is known at compile time.
METHOD 1:
One way of dealing with the problem is through the use of the typedef keyword. To

allocate a 2
dimensional array of integers recall that the following two notations result in the same

object code
being generated:
multi[row][col] = 1; *(*(multi + row) + col) = 1;
It is also true that the following two notations generate the same code:
multi[row] *(multi + row)
Since the one on the right must evaluate to a pointer, the array notation on the left

must also evaluate
to a pointer. In fact multi[0] will return a pointer to the first integer in the first

row, multi[1] a
pointer to the first integer of the second row, etc. Actually, multi[n] evaluates to a

pointer to that
array of integers that make up the n-th row of our 2 dimensional array. That is, multi

can be thought
of as an array of arrays and multi[n] as a pointer to the n-th array of this array of

arrays. Here the
word pointer is being used to represent an address value. While such usage is common in

the
literature, when reading such statements one must be careful to distinguish between the

constant
address of an array and a variable pointer which is a data object in itself.
Consider now:
--------------- Program 9.1 --------------------------------
/* Program 9.1 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <stdlib.h>
#define COLS 5
typedef int RowArray[COLS];
RowArray *rptr;
int main(void)
{
int nrows = 10;
int row, col;
rptr = malloc(nrows * COLS * sizeof(int));
for (row = 0; row < nrows; row++)
{
for (col = 0; col < COLS; col++)
{
rptr[row][col] = 17;
}
}
return 0;
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}
------------- End of Prog. 9.1 --------------------------------
Here I have assumed an ANSI compiler so a cast on the void pointer returned by malloc()

is not
required. If you are using an older K&R compiler you will have to cast using:
rptr = (RowArray *)malloc(.... etc.
Using this approach, rptr has all the characteristics of an array name name, (except that

rptr is
modifiable), and array notation may be used throughout the rest of the program. That also

means
that if you intend to write a function to modify the array contents, you must use COLS as

a part of
the formal parameter in that function, just as we did when discussing the passing of two

dimensional
arrays to a function.
METHOD 2:
In the METHOD 1 above, rptr turned out to be a pointer to type "one dimensional array of

COLS
integers". It turns out that there is syntax which can be used for this type without the

need of
typedef. If we write:
int (*xptr)[COLS];
the variable xptr will have all the same characteristics as the variable rptr in METHOD 1

above,
and we need not use the typedef keyword. Here xptr is a pointer to an array of integers

and the size
of that array is given by the #defined COLS. The parenthesis placement makes the pointer

notation
predominate, even though the array notation has higher precedence. i.e. had we written
int *xptr[COLS];
we would have defined xptr as an array of pointers holding the number of pointers equal

to that
#defined by COLS. That is not the same thing at all. However, arrays of pointers have

their use in
the dynamic allocation of two dimensional arrays, as will be seen in the next 2 methods.
METHOD 3:
Consider the case where we do not know the number of elements in each row at compile

time, i.e.
both the number of rows and number of columns must be determined at run time. One way of

doing
this would be to create an array of pointers to type int and then allocate space for each

row and point
these pointers at each row. Consider:
-------------- Program 9.2 ------------------------------------
/* Program 9.2 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <stdlib.h>
int main(void)
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{
int nrows = 5; /* Both nrows and ncols could be evaluated */
int ncols = 10; /* or read in at run time */
int row;
int **rowptr;
rowptr = malloc(nrows * sizeof(int *));
if (rowptr == NULL)
{
puts("
Failure to allocate room for row pointers.
");
exit(0);
}
printf("
Index Pointer(hex) Pointer(dec) Diff.(dec)");
for (row = 0; row < nrows; row++)
{
rowptr[row] = malloc(ncols * sizeof(int));
if (rowptr[row] == NULL)
{
printf("
Failure to allocate for row[%d]
",row);
exit(0);
}
printf("
%d %p %d", row, rowptr[row], rowptr[row]);
if (row > 0)
printf(" %d",(int)(rowptr[row] - rowptr[row-1]));
}
return 0;
}
--------------- End 9.2 ------------------------------------
In the above code rowptr is a pointer to pointer to type int. In this case it points to

the first element
of an array of pointers to type int. Consider the number of calls to malloc():
To get the array of pointers 1 call
To get space for the rows 5 calls
-----
Total 6 calls
If you choose to use this approach note that while you can use the array notation to

access individual
elements of the array, e.g. rowptr[row][col] = 17;, it does not mean that the data in the

"two
dimensional array" is contiguous in memory.
You can, however, use the array notation just as if it were a continuous block of memory.

For
example, you can write:
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rowptr[row][col] = 176;
just as if rowptr were the name of a two dimensional array created at compile time. Of

course row
and col must be within the bounds of the array you have created, just as with an array

created at
compile time.
If you want to have a contiguous block of memory dedicated to the storage of the elements

in the
array you can do it as follows:
METHOD 4:
In this method we allocate a block of memory to hold the whole array first. We then

create an array
of pointers to point to each row. Thus even though the array of pointers is being used,

the actual
array in memory is contiguous. The code looks like this:
----------------- Program 9.3 -----------------------------------
/* Program 9.3 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int **rptr;
int *aptr;
int *testptr;
int k;
int nrows = 5; /* Both nrows and ncols could be evaluated */
int ncols = 8; /* or read in at run time */
int row, col;
/* we now allocate the memory for the array */
aptr = malloc(nrows * ncols * sizeof(int));
if (aptr == NULL)
{
puts("
Failure to allocate room for the array");
exit(0);
}
/* next we allocate room for the pointers to the rows */
rptr = malloc(nrows * sizeof(int *));
if (rptr == NULL)
{
puts("
Failure to allocate room for pointers");
exit(0);
}
/* and now we 'point' the pointers */
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for (k = 0; k < nrows; k++)
{
rptr[k] = aptr + (k * ncols);
}
/* Now we illustrate how the row pointers are incremented */
printf("
Illustrating how row pointers are incremented");
printf("
Index Pointer(hex) Diff.(dec)");
for (row = 0; row < nrows; row++)
{
printf("
%d %p", row, rptr[row]);
if (row > 0)
printf(" %d",(rptr[row] - rptr[row-1]));
}
printf("
And now we print out the array
");
for (row = 0; row < nrows; row++)
{
for (col = 0; col < ncols; col++)
{
rptr[row][col] = row + col;
printf("%d ", rptr[row][col]);
}
putchar('
');
}
puts("
");
/* and here we illustrate that we are, in fact, dealing with
a 2 dimensional array in a contiguous block of memory. */
printf("And now we demonstrate that they are contiguous in memory
");
testptr = aptr;
for (row = 0; row < nrows; row++)
{
for (col = 0; col < ncols; col++)
{
printf("%d ", *(testptr++));
}
putchar('
');
}
return 0;
}
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------------- End Program 9.3 -----------------
Consider again, the number of calls to malloc()
To get room for the array itself 1 call
To get room for the array of ptrs 1 call
----
Total 2 calls
Now, each call to malloc() creates additional space overhead since malloc() is generally
implemented by the operating system forming a linked list which contains data concerning

the size
of the block. But, more importantly, with large arrays (several hundred rows) keeping

track of what
needs to be freed when the time comes can be more cumbersome. This, combined with the
contiguousness of the data block that permits initialization to all zeroes using memset()

would seem
to make the second alternative the preferred one.
As a final example on multidimensional arrays we will illustrate the dynamic allocation

of a three
dimensional array. This example will illustrate one more thing to watch when doing this

kind of
allocation. For reasons cited above we will use the approach outlined in alternative two.

Consider
the following code:
------------------- Program 9.4 -------------------------------------
/* Program 9.4 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <stdlib.h>
#include <stddef.h>
int X_DIM=16;
int Y_DIM=5;
int Z_DIM=3;
int main(void)
{
char *space;
char ***Arr3D;
int y, z;
ptrdiff_t diff;
/* first we set aside space for the array itself */
space = malloc(X_DIM * Y_DIM * Z_DIM * sizeof(char));
/* next we allocate space of an array of pointers, each
to eventually point to the first element of a
2 dimensional array of pointers to pointers */
Arr3D = malloc(Z_DIM * sizeof(char **));
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/* and for each of these we assign a pointer to a newly
allocated array of pointers to a row */
for (z = 0; z < Z_DIM; z++)
{
Arr3D[z] = malloc(Y_DIM * sizeof(char *));
/* and for each space in this array we put a pointer to
the first element of each row in the array space
originally allocated */
for (y = 0; y < Y_DIM; y++)
{
Arr3D[z][y] = space + (z*(X_DIM * Y_DIM) + y*X_DIM);
}
}
/* And, now we check each address in our 3D array to see if
the indexing of the Arr3d pointer leads through in a
continuous manner */
for (z = 0; z < Z_DIM; z++)
{
printf("Location of array %d is %p
", z, *Arr3D[z]);
for ( y = 0; y < Y_DIM; y++)
{
printf(" Array %d and Row %d starts at %p", z, y, Arr3D[z][y]);
diff = Arr3D[z][y] - space;
printf(" diff = %d ",diff);
printf(" z = %d y = %d
", z, y);
}
}
return 0;
}
------------------- End of Prog. 9.4 ----------------------------
If you have followed this tutorial up to this point you should have no problem

deciphering the above
on the basis of the comments alone. There are a couple of points that should be made

however. Let's
start with the line which reads:
Arr3D[z][y] = space + (z*(X_DIM * Y_DIM) + y*X_DIM);
Note that here space is a character pointer, which is the same type as Arr3D[z][y]. It is

important
that when adding an integer, such as that obtained by evaluation of the expression (z*

(X_DIM *
Y_DIM) + y*X_DIM), to a pointer, the result is a new pointer value. And when assigning

pointer
values to pointer variables the data types of the value and variable must match.
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Pointers to Functions
Up to this point we have been discussing pointers to data objects. C also permits the

declaration of
pointers to functions. Pointers to functions have a variety of uses and some of them will

be
discussed here.
Consider the following real problem. You want to write a function that is capable of

sorting virtually
any collection of data that can be stored in an array. This might be an array of strings,

or integers, or
floats, or even structures. The sorting algorithm can be the same for all. For example,

it could be a
simple bubble sort algorithm, or the more complex shell or quick sort algorithm. We'll

use a simple
bubble sort for demonstration purposes.
Sedgewick [1] has described the bubble sort using C code by setting up a function which

when
passed a pointer to the array would sort it. If we call that function bubble(), a sort

program is
described by bubble_1.c, which follows:
/*-------------------- bubble_1.c --------------------*/
/* Program bubble_1.c from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
int arr[10] = { 3,6,1,2,3,8,4,1,7,2};
void bubble(int a[], int N);
int main(void)
{
int i;
putchar('
');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('
');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
return 0;
}
void bubble(int a[], int N)
{
int i, j, t;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
if (a[j-1] > a[j])
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{
t = a[j-1];
a[j-1] = a[j];
a[j] = t;
}
}
}
}
/*---------------------- end bubble_1.c -----------------------*/
The bubble sort is one of the simpler sorts. The algorithm scans the array from the

second to the last
element comparing each element with the one which precedes it. If the one that precedes

it is larger
than the current element, the two are swapped so the larger one is closer to the end of

the array. On
the first pass, this results in the largest element ending up at the end of the array.

The array is now
limited to all elements except the last and the process repeated. This puts the next

largest element at
a point preceding the largest element. The process is repeated for a number of times

equal to the
number of elements minus 1. The end result is a sorted array.
Here our function is designed to sort an array of integers. Thus in line 1 we are

comparing integers
and in lines 2 through 4 we are using temporary integer storage to store integers. What

we want to
do now is see if we can convert this code so we can use any data type, i.e. not be

restricted to
integers.
At the same time we don't want to have to analyze our algorithm and the code associated

with it
each time we use it. We start by removing the comparison from within the function bubble

() so as
to make it relatively easy to modify the comparison function without having to re-write

portions
related to the actual algorithm. This results in bubble_2.c:
/*---------------------- bubble_2.c -------------------------*/
/* Program bubble_2.c from PTRTUT10.HTM 6/13/97 */
/* Separating the comparison function */
#include <stdio.h>
int arr[10] = { 3,6,1,2,3,8,4,1,7,2};
void bubble(int a[], int N);
int compare(int m, int n);
int main(void)
{
int i;
putchar('
');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
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putchar('
');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
return 0;
}
void bubble(int a[], int N)
{
int i, j, t;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
if (compare(a[j-1], a[j]))
{
t = a[j-1];
a[j-1] = a[j];
a[j] = t;
}
}
}
}
int compare(int m, int n)
{
return (m > n);
}
/*--------------------- end of bubble_2.c -----------------------*/
If our goal is to make our sort routine data type independent, one way of doing this is

to use pointers
to type void to point to the data instead of using the integer data type. As a start in

that direction let's
modify a few things in the above so that pointers can be used. To begin with, we'll stick

with
pointers to type integer.
/*----------------------- bubble_3.c -------------------------*/
/* Program bubble_3.c from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
int arr[10] = { 3,6,1,2,3,8,4,1,7,2};
void bubble(int *p, int N);
int compare(int *m, int *n);
int main(void)
{
int i;
putchar('
');
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for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('
');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
return 0;
}
void bubble(int *p, int N)
{
int i, j, t;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
if (compare(&p[j-1], &p[j]))
{
t = p[j-1];
p[j-1] = p[j];
p[j] = t;
}
}
}
}
int compare(int *m, int *n)
{
return (*m > *n);
}
/*------------------ end of bubble3.c -------------------------*/
Note the changes. We are now passing a pointer to an integer (or array of integers) to

bubble(). And
from within bubble we are passing pointers to the elements of the array that we want to

compare to
our comparison function. And, of course we are dereferencing these pointer in our

compare()
function in order to make the actual comparison. Our next step will be to convert the

pointers in
bubble() to pointers to type void so that that function will become more type

insensitive. This is
shown in bubble_4.
/*------------------ bubble_4.c ----------------------------*/
/* Program bubble_4.c from PTRTUT10,HTM 6/13/97 */
#include <stdio.h>
int arr[10] = { 3,6,1,2,3,8,4,1,7,2};
void bubble(int *p, int N);
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int compare(void *m, void *n);
int main(void)
{
int i;
putchar('
');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('
');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
return 0;
}
void bubble(int *p, int N)
{
int i, j, t;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
if (compare((void *)&p[j-1], (void *)&p[j]))
{
t = p[j-1];
p[j-1] = p[j];
p[j] = t;
}
}
}
}
int compare(void *m, void *n)
{
int *m1, *n1;
m1 = (int *)m;
n1 = (int *)n;
return (*m1 > *n1);
}
/*------------------ end of bubble_4.c ---------------------*/
Note that, in doing this, in compare() we had to introduce the casting of the void

pointer types
passed to the actual type being sorted. But, as we'll see later that's okay. And since

what is being
passed to bubble() is still a pointer to an array of integers, we had to cast these

pointers to void
pointers when we passed them as parameters in our call to compare().
We now address the problem of what we pass to bubble(). We want to make the first

parameter of
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that function a void pointer also. But, that means that within bubble() we need to do

something
about the variable t, which is currently an integer. Also, where we use t = p[j-1]; the

type of p[j-1]
needs to be known in order to know how many bytes to copy to the variable t (or whatever

we
replace t with).
Currently, in bubble_4.c, knowledge within bubble() as to the type of the data being

sorted (and
hence the size of each individual element) is obtained from the fact that the first

parameter is a
pointer to type integer. If we are going to be able to use bubble() to sort any type of

data, we need to
make that pointer a pointer to type void. But, in doing so we are going to lose

information
concerning the size of individual elements within the array. So, in bubble_5.c we will

add a separate
parameter to handle this size information.
These changes, from bubble4.c to bubble5.c are, perhaps, a bit more extensive than those

we have
made in the past. So, compare the two modules carefully for differences.
/*---------------------- bubble5.c ---------------------------*/
/* Program bubble_5.c from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <string.h>
long arr[10] = { 3,6,1,2,3,8,4,1,7,2};
void bubble(void *p, size_t width, int N);
int compare(void *m, void *n);
int main(void)
{
int i;
putchar('
');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr, sizeof(long), 10);
putchar('
');
for (i = 0; i < 10; i++)
{
printf("%ld ", arr[i]);
}
return 0;
}
void bubble(void *p, size_t width, int N)
{
int i, j;
unsigned char buf[4];
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unsigned char *bp = p;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
if (compare((void *)(bp + width*(j-1)),
(void *)(bp + j*width))) /* 1 */
{
/* t = p[j-1]; */
memcpy(buf, bp + width*(j-1), width);
/* p[j-1] = p[j]; */
memcpy(bp + width*(j-1), bp + j*width , width);
/* p[j] = t; */
memcpy(bp + j*width, buf, width);
}
}
}
}
int compare(void *m, void *n)
{
long *m1, *n1;
m1 = (long *)m;
n1 = (long *)n;
return (*m1 > *n1);
}
/*--------------------- end of bubble5.c ---------------------*/
Note that I have changed the data type of the array from int to long to illustrate the

changes needed
in the compare() function. Within bubble() I've done away with the variable t (which we

would
have had to change from type int to type long). I have added a buffer of size 4 unsigned

characters,
which is the size needed to hold a long (this will change again in future modifications

to this code).
The unsigned character pointer *bp is used to point to the base of the array to be

sorted, i.e. to the
first element of that array.
We also had to modify what we passed to compare(), and how we do the swapping of elements

that
the comparison indicates need swapping. Use of memcpy() and pointer notation instead of

array
notation work towards this reduction in type sensitivity.
Again, making a careful comparison of bubble5.c with bubble4.c can result in improved
understanding of what is happening and why.
We move now to bubble6.c where we use the same function bubble() that we used in

bubble5.c to
sort strings instead of long integers. Of course we have to change the comparison

function since the
means by which strings are compared is different from that by which long integers are

compared.
And,in bubble6.c we have deleted the lines within bubble() that were commented out in

bubble5.c.
/*--------------------- bubble6.c ---------------------*/
/* Program bubble_6.c from PTRTUT10.HTM 6/13/97 */
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#include <stdio.h>
#include <string.h>
#define MAX_BUF 256
char arr2[5][20] = { "Mickey Mouse",
"Donald Duck",
"Minnie Mouse",
"Goofy",
"Ted Jensen" };
void bubble(void *p, int width, int N);
int compare(void *m, void *n);
int main(void)
{
int i;
putchar('
');
for (i = 0; i < 5; i++)
{
printf("%s
", arr2[i]);
}
bubble(arr2, 20, 5);
putchar('
');
for (i = 0; i < 5; i++)
{
printf("%s
", arr2[i]);
}
return 0;
}
void bubble(void *p, int width, int N)
{
int i, j, k;
unsigned char buf[MAX_BUF];
unsigned char *bp = p;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
k = compare((void *)(bp + width*(j-1)), (void *)(bp + j*width));
if (k > 0)
{
memcpy(buf, bp + width*(j-1), width);
memcpy(bp + width*(j-1), bp + j*width , width);
memcpy(bp + j*width, buf, width);
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}
}
}
}
int compare(void *m, void *n)
{
char *m1 = m;
char *n1 = n;
return (strcmp(m1,n1));
}
/*------------------- end of bubble6.c ---------------------*/
But, the fact that bubble() was unchanged from that used in bubble5.c indicates that that

function is
capable of sorting a wide variety of data types. What is left to do is to pass to bubble

() the name of
the comparison function we want to use so that it can be truly universal. Just as the

name of an array
is the address of the first element of the array in the data segment, the name of a

function decays into
the address of that function in the code segment. Thus we need to use a pointer to a

function. In this
case the comparison function.
Pointers to functions must match the functions pointed to in the number and types of the

parameters
and the type of the return value. In our case, we declare our function pointer as:
int (*fptr)(const void *p1, const void *p2);
Note that were we to write:
int *fptr(const void *p1, const void *p2);
we would have a function prototype for a function which returned a pointer to type int.

That is
because in C the parenthesis () operator have a higher precedence than the pointer *

operator. By
putting the parenthesis around the string (*fptr) we indicate that we are declaring a

function pointer.
We now modify our declaration of bubble() by adding, as its 4th parameter, a function

pointer of
the proper type. It's function prototype becomes:
void bubble(void *p, int width, int N,
int(*fptr)(const void *, const void *));
When we call the bubble(), we insert the name of the comparison function that we want to

use.
bubble7.c illustrate how this approach permits the use of the same bubble() function for

sorting
different types of data.
/*------------------- bubble7.c ------------------*/
/* Program bubble_7.c from PTRTUT10.HTM 6/10/97 */
#include <stdio.h>
#include <string.h>
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#define MAX_BUF 256
long arr[10] = { 3,6,1,2,3,8,4,1,7,2};
char arr2[5][20] = { "Mickey Mouse",
"Donald Duck",
"Minnie Mouse",
"Goofy",
"Ted Jensen" };
void bubble(void *p, int width, int N,
int(*fptr)(const void *, const void *));
int compare_string(const void *m, const void *n);
int compare_long(const void *m, const void *n);
int main(void)
{
int i;
puts("
Before Sorting:
");
for (i = 0; i < 10; i++) /* show the long ints */
{
printf("%ld ",arr[i]);
}
puts("
");
for (i = 0; i < 5; i++) /* show the strings */
{
printf("%s
", arr2[i]);
}
bubble(arr, 4, 10, compare_long); /* sort the longs */
bubble(arr2, 20, 5, compare_string); /* sort the strings */
puts("
After Sorting:
");
for (i = 0; i < 10; i++) /* show the sorted longs */
{
printf("%d ",arr[i]);
}
puts("
");
for (i = 0; i < 5; i++) /* show the sorted strings */
{
printf("%s
", arr2[i]);
}
return 0;
}
void bubble(void *p, int width, int N,
int(*fptr)(const void *, const void *))
{
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int i, j, k;
unsigned char buf[MAX_BUF];
unsigned char *bp = p;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
k = fptr((void *)(bp + width*(j-1)), (void *)(bp + j*width));
if (k > 0)
{
memcpy(buf, bp + width*(j-1), width);
memcpy(bp + width*(j-1), bp + j*width , width);
memcpy(bp + j*width, buf, width);
}
}
}
}
int compare_string(const void *m, const void *n)
{
char *m1 = (char *)m;
char *n1 = (char *)n;
return (strcmp(m1,n1));
}
int compare_long(const void *m, const void *n)
{
long *m1, *n1;
m1 = (long *)m;
n1 = (long *)n;
return (*m1 > *n1);
}
/*----------------- end of bubble7.c -----------------*/
C Pointers Tutorial: Pointers to Functions Page 11 of 11
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