Declaration of Pointer with array and single double and Triple Pointer : Part 2

 

 

Declaration of Pointer: Part 2

The Two fundamental operators used with the pointers are:

1.       Address operator &

2.       Indirection operator *

 

#include<stdio.h>

void main()

{

Int a=5;

Int *p;   /*Pointer declaration*/

P=&a;  /*copying address of variable to the pointer p */

*p =10;  /* use of pointer to change the value of variable a*/

printf(“%d”, a);

printf(“%d”, *p);

printf(“%d”, *(&a));

}

All the printf statements in the above program will give the output as 10

 

Memory and Pointer

 

int *p1;

float *p2;

char *p3;

all pointers p1, p2, p3 get 2 bytes each.

 

Suppose if three pointers are declared for int, float, char. All the three pointers will occupy 2 bytes in the memory. This is because all the memory addresses are integer values ranging from 0 to 65536.

Example

                Void main()

{ int a=5, *p1;

Float b=2.5, *p2;

Char c=’a’, *p3;

p1=&a;

p2=&b;

p3=&c;

printf(“%d”, sizeof(p1));

printf(“%d”, sizeof(p2));

printf(“%d”,sizeof(p3));

}

Output 2 2 2

 

Pointers and Address:-

Void main(){

Int a=5;

Int *p;

P =&a;

Printf(“%p”,p);   //print the actual address of pointer//

Printf(“%d”,p);   //print the address in integer formata //

Printf(“%u”,p);   //print the address  of pointer p as unsigned integer //

Printf(“%X”, p);  // print the address of pointer p as hexadecimal //

}

                Output: fff4, -12, 65524, FFF4

                Let us evaluate the output the first answer using %p is print the actual address of pointer which is fff4. Second answer is a negative value this is because the range of pointers if from 0 to 65536 and variable are allocated memory from top to bottom i.e. from 65536 then 65535, 65534 and so on until 0.

Third answer is 65524 that means if we convert fff4 (hexadecimal) into decimal we get 65524 and difference of bottom to top (655524 – 65536 = -12) that’s why we will get second answer into integer.

The last answer is hexadecimal into capital %X which is FFF4.

 

Pointers Expression:-

Like any other variable, pointer variables can be used in expression.

1.       C allows us to add or subtract integers from pointers.

Ex.          Sum = sum + *p1;

Ex.          *p3 = *p1 + *p2

Ex.          Result = 50 - *p1;

Ex.          Ans = *p1 + *p2++;

 

2.       Two Pointers can not be multiplied and divided.

3.   Declaration with array:

  Void main()

{

Int a[5]={10,20,30,40,50};

Int *p;

p=a  //address of array can be pointed into the pointer without the use of & operator //

printf(“%d”,*p);   //10

printf(“%d”,*(p+1));  //20

printf(“%d”, *(p+2)); //30

printf(“%d”, *(p+3));   //40

printf(“%d”, *(p+4)); //50

getch();

}

 Pointers to Pointers:

Pointers store the address of a variable, similarly the address of a pointer can also be stored in some other pointer.

Void main()

{

Int a=5;

Int *p1;  // pointer to an integer :single pointer

Int **p2 //  pointer to pointer to an integer:  Double pointer

Int ***p3  // pointer to pointer to pointer to an integer : Triple pointer

P1=&a;

P2=&p1;

P3=&p2;

Printf(“%d”,a) // 5

Printf(“%d”,*p1)  //5

Printf(“%d”,**p2) //5

Printf(“%d”, ***p3) //5

 

}