Revision as of 15:37, 15 June 2014

Pointer as the word implies just points. It is a datatype in C which is used to hold memory address. Pointers may point to any datatype in C and based on that pointer arithmetic is done.

A pointer is declared using * as follows:

int* p; //p is a pointer to int. int *p is also syntactically correct but * here is always taken with int as int* and p is the name for the variable. Many take this as *p which is not correct
char* y; //y is a pointer to char

Actually pointer operation is very simple as there are just two operators for using pointer.

• p -> gives the content of the location in p.

What is the content depends on the type of p. If p is an integer pointer, *p will return 4 bytes from the location contained in p. If p is a char pointer *p will return 1 byte from the location in p.

int *p, a;
scanf("%d",&a); //memory address of a is passed to scanf and it stores the keyboard entered value in that location
p = &a; //Memory address of a is stored in p
printf("*p = %d", *p); //Content of p is printed by printf. Since %d is used, 4 bytes (assuming sizeof int is 4) from the memory location given by p is converted to integer and printed.

int *p, a;
p = &a;
scanf("%d",p);
printf("a = %d *p = %d", a, *p);

Exercise Questions

<quiz display="simple"> {What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int a = 5;
 int* p = &a;
 printf("%d", ++*p);

} </syntaxhighlight> |type="{}" /} { 6 } ||p is pointing to the address of a. *p will return the content of a which is 5 and ++ will increment it to 6.

{What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 char a[] = "Hello World";
 char* p = &a;
 printf("%s", p+2 );

} </syntaxhighlight> |type="()" /} -Compiler Error -World +llo World -Runtime Error

||Since p is a char pointer p+2 will add 2 to p (since sizeof(char) is 1). So, p+2 will be pointing to the string "llo World".

{What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int a;
 int* p = &a;
 printf("%zu", sizeof( *(char*)p  ));

} </syntaxhighlight> |type="()" /}

+1 -2 -4 -Compile Error

||p is typecasted to char pointer and then dereferenced. So, returned type will be char and sizeof(char) is 1.

{Is the following code legal? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int a = 1;
 if((char*)  &a)
 {
   printf("My machine is little endian");
 }
 else
 {
    printf("My machine is big endian\n");
 }

} </syntaxhighlight> |type="()" /} +Yes -No ||On a little endian machine the lower address will contain the least significant byte. Suppose a is stored at address 1000 and contains 1, then character at 1000 will be 1, if the machine is little endian.

{What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int *a = (int*) 1;
 printf("%d", a);

} </syntaxhighlight> |type="()" /} -Garbage value +1 -0 -Compile error -Segmentation fault ||Assigning int values to pointer variable is possible. Only when we dereference the variable using *, we get a segmentation fault.

{What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int a = 1, *p, **pp;
 p = &a;
 pp = p;
 printf("%d", **pp);

} </syntaxhighlight> |type="()" /} -Garbage value -1 -Compile error +Segmentation fault ||Here, p is having address of a and the same is copied to pp. So, *pp will give 1 which is contained in a, but **p will use 1 as an address and tries to access the memory location 1, giving segmentation fault.

{What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int a = 1, *p, **pp;
 p = &a;
 pp = p;
 printf("%d", *pp);

} </syntaxhighlight> |type="()" /} +1 -Garbage value -Compile error -Segmentation fault ||Here, p is having address of a and the same is copied to pp. So, *pp will give 1 which is contained in a.

{Assuming a little endian machine, what will be the output of the following program? <syntaxhighlight lang="c">

1. include<stdio.h>

fun(int a) {

 char *arr[] = {"0000", "0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};
 unsigned char* p = (unsigned char*) &a ;
 p+=3;
 int i;
 for(i = 0; i < sizeof a; i++)
 {
   int d = (*p)>>4; 
   printf("%s", arr[d]);
   d = (*p) & 0xf;
   printf("%s ", arr[d]);
   p--;
 }

}

int main() {

int a;
scanf("%d", &a);
fun(a);

} </syntaxhighlight> |type="()" /} +Print the binary of the input number -Compile error -Runtime error -Compiler dependent output ||The code is printing the binary equivalent of the input number. Suppose a is stored starting from address 1000. Since, we assume a little endian machine, the LSB of a will be stored at address 1000 and MSB will be stored at address 1003. So, we make a char pointer to 1003 and take out the MSB. Then using shift operator we get the most significant 4 bits from it and then the lest significant 4 bits. We repeat the same for the other three bytes of a. </quiz>