Chapter 10

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

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('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('\n');

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])
{
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

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('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('\n');

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

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('\n');

for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('\n');

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

int arr[10] = { 3,6,1,2,3,8,4,1,7,2};

void bubble(int *p, int N);
int compare(void *m, void *n);

int main(void)
{
int i;
putchar('\n');

for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('\n');

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 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
#include

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('\n');

for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr, sizeof(long), 10);
putchar('\n');

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];
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 */

#include
#include

#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('\n');

for (i = 0; i < 5; i++)
{
printf("%s\n", arr2[i]);
}
bubble(arr2, 20, 5);
putchar('\n\n');

for (i = 0; i < 5; i++)
{
printf("%s\n", 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);
}
}
}
}

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
#include

#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("\nBefore Sorting:\n");

for (i = 0; i < 10; i++) /* show the long ints */
{
printf("%ld ",arr[i]);
}
puts("\n");

for (i = 0; i < 5; i++) /* show the strings */
{
printf("%s\n", arr2[i]);
}
bubble(arr, 4, 10, compare_long); /* sort the longs */
bubble(arr2, 20, 5, compare_string); /* sort the strings */
puts("\n\nAfter Sorting:\n");

for (i = 0; i < 10; i++) /* show the sorted longs */
{
printf("%d ",arr[i]);
}
puts("\n");

for (i = 0; i < 5; i++) /* show the sorted strings */
{
printf("%s\n", arr2[i]);
}
return 0;
}

void bubble(void *p, int width, int N,
int(*fptr)(const void *, const void *))
{
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);
}

Chapter 8

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 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.
Continue with Pointer Tutorial
Back to Table of Contents

Chapter 7

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. Here the arithmetic being used
is of a special kind called "pointer arithmetic" is being used. 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:
The address of the first element of the array, which is returned by the
expression multi, i.e., the name of the array.
The size of the type of the elements of the array, in this case sizeof(int).
The 2nd dimension of the array
The specific index value for the first dimension, row in this case.
The specific index value for the second dimension, col in this case.
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 in the 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 multi-dimensional arrays. That is if we are talking about 3
dimensional arrays, the 2nd and 3rd dimension must be specified in the parameter
definition.
Continue with Pointer Tutorial
Back to Table of Contents

Chapter 6

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 '\0').
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"
.
.
}

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
^
|_____ starting at the address &multi[0][0]


Note that I did not write multi[0] = "0123456789". Had I done so a terminating
'\0' would have been implied since whenever double quotes are used a '\0'
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
#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("\n%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).
Continue with Pointer Tutorial
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Chapter 5

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
#include

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("\n%s ",my_struct.fname);
printf("%s\n",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 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:
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
#include

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("\n%s ",my_struct.fname);
printf("%s\n",my_struct.lname);
my_struct.age = 63;
show_name(st_ptr); /* pass the pointer */
return 0;
}

void show_name(struct tag *p)
{
printf("\n%s ", p->fname); /* p points to a structure */
printf("%s ", p->lname);
printf("%d\n", 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.

Continue with Pointer Tutorial
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Chapter 4

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] != '\0')
{
dest[i] = source[i];
i++;
}
dest[i] = '\0';
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
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 != '\0')

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.
Continue with Pointer Tutorial
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CHAPTER 3

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 '\0'). 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] = '\0';


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 '\0'. 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', '\0',};

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 ( '\0' ) 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\0.
Now, consider the following program:
------------------program 3.1-------------------------------------

/* Program 3.1 from PTRTUT10.HTM 6/13/97 */

#include

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('\n'); /* move down one line on the screen */
while(*pA != '\0') /* line A (see text) */
{
*pB++ = *pA++; /* line B (see text) */
}
*pB = '\0'; /* 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 '\0'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 '\0'), 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.
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 != '\0')
{
*p++ = *source++;
}
*p = '\0';
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 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 '\0'. 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.
Continue with Pointer Tutorial
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Chapter 2

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, 2 bytes would be
copied, if a long, 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 if ten integers in a
row. That is, 20 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 2 to ptr instead of 1, so the pointer "points to" the next
integer, at memory location 102. Similarly, were the ptr declared as a pointer
to a long, it would add 4 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. 2 for an integer, 4 for a long, etc.).
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
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

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("\n\n");
for (i = 0; i < 6; i++)
{
printf("my_array[%d] = %d ",i,my_array[i]); /*<-- A */
printf("ptr + %d = %d\n",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\n",i, *ptr++);

and run it again... then change it to:
printf("ptr + %d = %d\n",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:
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.
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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Chapter 1

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 PC's the size of an integer variable is 2 bytes, and that of a long
integer is 4 bytes. In C the size of a variable type such as an integer need not
be the same on all types of machines.
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 2 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 2 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".
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 2 byte integers so all copying of rvalues from one storage location to the other is done by copying 2 bytes. Had we been using long integers, we would be copying 4 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. Some computers, such as the IBM PC might require special handling to hold a segment and offset under certain circumstances. 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; 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\n",*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

int j, k;
int *ptr;

int main(void)
{
j = 1;
k = 2;
ptr = &k;
printf("\n");
printf("j has the value %d and is stored at %p\n", j, (void *)&j);
printf("k has the value %d and is stored at %p\n", k, (void *)&k);
printf("ptr has the value %p and is stored at %p\n", ptr, (void *)&ptr);
printf("The value of the integer pointed to by ptr is %d\n", *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).
References for Chapter 1:
"The C Programming Language" 2nd Edition
B. Kernighan and D. Ritchie
Prentice Hall
ISBN 0-13-110362-8

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