### Pointers in c

Consider the declaration,

int i = 3 ;

This declaration tells the C compiler to:

1. Reserve space in memory to hold the integer value.

2. Associate the name i with this memory location.

3. Store the value 3 at this location.

We may represent i 's location in the memory by the following memory map:

Figure 1

We see that the computer has selected memory location 6485 as the place to store the value 3. This location number 6485 is not a number to be relied upon, because some other time the computer may choose a different location for storing the value 3.

We can print this address through the following statement :
printf ( "\nAddress of i = %u", &i ) ; /* o/p : 6485 */

Look at the printf( ) statement carefully. '&' used in this statement is C's 'address of' operator. The expression &i returns the address of the variable i, which in this case happens to be 6485.

The other pointer operator available in C is '*', called 'value at address' operator. It returns the value stored at a particular address. The 'value at address' operator is also called 'indirection' operator.

Observe carefully the following statements :

printf ( "\nAddress of i = %u", &i ) ; /* o/p : 6485 */

printf ( "\nValue of i = %d", i ) ; /* o/p : 3 */

printf ( "\nValue of i = %d", *( &i ) ) ; /* o/p : 3 */

Note that printing the value of *( &i ) is same as printing the value of i.

Let us now see what are pointers and how they can be used in various expressions. We have seen in the above section that the expression &i returns the address of i. If we so desire this address can be collected in a variable by saying,

j = &i ;

But remember that j is not an ordinary variable like any other integer variable. It is a variable which contains the address of another variable ( i in this case ).

Since j is a variable, the compiler must provide it space in memory. The following memory map illustrates the contents of i and j.

Figure 2

As you can see, i 's value is 3 and j 's value is i 's address.

But, we can't use j in a program without declaring it. And since j is a variable which contains the address of i, it is declared as,

int *j ;

This declaration tells the compiler that j will be used to store the address of an integer value - in other words j points to an integer.

How do we justify the usage of * in the declaration,

int *j ;

Let us go by the meaning of *. It stands for 'value at address'. Thus, int *j would mean:

1. value at address stored in j is an int

2. j contains address of an int

3. j points to an int

4. j is a pointer which points in the direction of an int

So we can conclude that pointer is a variable which contains address of another variable.

More pointer types
The way we have used an int pointer in the above example we can also build a char pointer, a float pointer, a long int pointer, etc. This is shown below:

char ch = 'A' ;

float a = 3.14 ;

long int j = 40000L ;

char *dh ;

float *b ;

long int *k ;

dh = &ch ;

b = &a ;

k = &j ;

Pointer to Pointer
If we store address of an integer pointer in a variable that variable becomes pointer to an integer pointer. And if we store address of a pointer to an integer pointer in a variable that becomes a pointer to a pointer to an integer pointer. This is shown in the following code segment:

int i = 10 ;

int *j ; // integer pointer

int **k ; // pointer to an integer pointer

int ***l ; // pointer to a pointer to an integer pointer

j = &i ;

k = &j ;

l = &k ;

// print 10 using i, j, k, l

printf ( "%d %d %d %d", i, *j, **k, ***l ) ;

Pointer Arithmetic
Following operations can be performed on a pointer:

1. Addition of a number to a pointer. For example,

int i = 4, *j, *k ;

j = &i ;

j = j + 1 ;

j = j + 9 ;

k = j + 3 ;

1. Subtraction of a number from a pointer. For example,

int i = 4, *j, *k ;

j = &i ;

j = j - 2 ;

j = j - 5 ;

k = j - 6 ;

1. Subtraction of a pointer from a pointer. For example,

int i = 4, j = 5, *p, *q, d ;

p = &i ;

q = &j ;

d = q - p ;

The following program catches the essence of all that we have said about pointers so far.

main( )

{

float a = 3.14, *b ;

char ch = 'z', *dh ;

int i = 25, *j ;

printf ( "%u%u%u%u", &a, &ch, &i ) ; /* prints addresses of a, ch and i */

b = &a ; /* assigns address of a to b */

dh = &ch ; /* assigns address of ch to dh */

j = &i ; /* assigns address of i to j */

printf ( "%u%u%u", *b, *dh, *j ) ; /* prints value at address stored in b,

dh and j respectively */

printf ( "%u%u%u%u", b, dh,j ) ; /* prints the value in b, dh, j which is

the address of a, ch, j respectively*/

b++ ; /* makes the pointer b point to a location adjacent to variable a */

dh++ ; /* makes the pointer dh point to a location adjacent to variable ch */

j++ ;/* makes the pointer j point to the location adjacent to variable i */

/* += -= operators can be used in pointer arithmetic */

b+= 3 ; dh += 8 ;

j-= 3 ;

printf ( "%u%u%u%u", b, dh, j );

}

near, far and huge Pointer
(The information that follows is specific to DOS operating system only)

While working under DOS only 1 mb (10,48,580 bytes) of memory is accessible. Any of these memory locations are accessed using CPU registers. Under DOS the CPU registers are only 16 bits long. Therefore, the minimum value present in a CPU register could be 0, and maximum 65,535. Then how do we access memory locations beyond 65535th byte? By using two registers (segment and offset) in conjunction. For this the total memory (1 mb) is divided into a number of units each comprising 65,536 (64 kb) locations. Each such unit is called a segment. Each segment always begins at a location number which is exactly divisible by 16. The segment register contains the address where a segment begins, whereas the offset register contains the offset of the data/code from where the segment begins. For example, let us consider the first byte in B block of video memory. The segment address of video memory is B0000h (20-bit address), whereas the offset value of the first byte in the upper 32K block of this segment is 8000h. Figure 3.

Since 8000h is a 16-bit address it can be easily placed in the offset register, but how do we store the 20-bit address B0000h in a 16-bit segment register? For this out of B0000h only first four hex digits (16 bits) are stored in segment register. We can afford to do this because a segment address is always a multiple of 16 and hence always contains a 0 as the last digit. Therefore, the first byte in the upper 32K chunk of B block of video memory is referred using segment:offset format as B000h:8000h. Thus, the offset register works relative to segment register. Using both these, we can point to a specific location anywhere in the 1 mb address space.

Suppose we want to write a character `A' at location B000:8000. We must convert this address into a form which C understands. This is done by simply writing the segment and offset addresses side by side to obtain a 32 bit address. In our example this address would be 0xB0008000. Now whether C would support this 32 bit address or not depends upon the memory model in use. For example, if we are using a large data model (compact, large, huge) the above address is acceptable. This is because in these models all pointers to data are 32 bits long. As against this, if we are using a small data model (tiny, small, medium) the above address won't work since in these models each pointer is 16 bits long.

What if we are working in small data model and still want to access the first byte of the upper 32K chunk of B block of video memory? In such cases both Microsoft C and Turbo C provide a keyword called far, which is used as shown below,

char far *s = 0XB0008000;

A far pointer is always treated as 32 bit pointer and contains both a segment address and an offset.

A huge pointer is also 32 bits long, again containing a segment address and an offset. However, there are a few differences between a far pointer and a huge pointer.

A near pointer is only 16 bits long, it uses the contents of CS register (if the pointer is pointing to code) or contents of DS register (if the pointer is pointing to data) for the segment part, whereas the offset part is stored in the 16-bit near pointer. Using near pointer limits your data/code to current 64 kb segment.

The following table captures the essence of these different types of pointers along with the pointer type supported by each memory model.

Memory model Code Pointer Data Pointer Code size Data size
Tiny near near <--------------- 64 KB ------------->
Small near near 64 KB 64 KB
Medium far near 1 MB 64 KB
Compact near far 64 KB 1 MB
Large far far 1 MB 1 MB
Huge far huge 1 MB *
* single data element (say an array) can occupy more than 1 segment.

A far pointer (32 bit) contains the segment as well as the offset. By using far pointers we can have multiple code segments, which in turn allow you to have programs longer than 64 kb. Likewise, with far data pointers we can address more than 64 kb worth of data. However, while using far pointers some problems may crop up as is illustrated by the following program.

main( )

{

char far *a = OX00000120;

char far *b = OX00100020;

char far *c = OX00120000;

if ( a == b )

printf ( "Hello" ) ;

if ( a == c )

printf ( "Hi" ) ;

if ( b == c )

printf ( "Hello Hi" ) ;

if ( a > b && a > c && b > c )

printf ( "Bye" ) ;

}

Note that all the 32 bit addresses stored in variables a, b, and c refer to the same memory location. This deduces from the method of obtaining the 20-bit physical address from the segment:offset pair. This is shown below.

00000 segment address left shifted by 4 bits

--------

00100 segment address left shifted by 4 bits

--------

00120 segment address left shifted by 4 bits

--------

Now if a, b and c refer to same location in memory we expect the first three ifs to be satisfied. However this doesn't happen. This is because while comparing the far pointers using == (and !=) the full 32-bit value is used and since the 32-bit values are different the ifs fail. The last if however gets satisfied, because while comparing using > (and >=, <, <= ) only the offset value is used for comparison. And the offset values of a, b and c are such that the last condition is satisfied.

These limitations are overcome if we use huge pointer instead of far pointers. Unlike far pointers huge pointers are `normalized' to avoid these problems. What is a normalized pointer? It is a 32- bit pointer which has as much of its value in the segment address as possible. Since a segment can start every 16 bytes, this means that the offset will only have a value from 0 to F.

How do we normalize a pointer? Simple. Convert it to its 20-bit address then use the the left 16 bits for the segment address and the right 4 bits for the offset address. For example, given the pointer 500D:9407, we convert it to the absolute address 594D7, which we then normalize to 594D:0007.

huge pointers are always kept normalized. As a result, for any given memory address there is only one possible huge address - segment:offset pair for it. Run the above program using huge instead of far and now you would find that the first three ifs are satisfied, whereas the fourth fails. This is more logical than the result obtained while using far.

But then there is a price to be paid for using huge pointers. Huge pointer arithmetic is done with calls to special subroutines. Because of this, huge pointer arithmetic is significantly slower than that of far or near pointers.

Tips:

1. Addresses must always be printed using %u or %p.

1. If %p is used address is printed in segment:Offset form. If %u is used only offset address is printed (specific to DOS).

1. In DOS there are three types of pointers :

near (2 bytes), far (4 bytes) and huge (4 bytes). >

In Unix and Windows every pointer is a 4 byte entity.

1. A pointer when incremented always points to an immediately next location of its type.

1. The only legal pointer arithmetic is :

pointer + number,

pointer - number,

pointer - pointer.

1. Don't attempt the following arithmetic operations on pointer. They won't work :