1. Chapter 5
POINTERs

2. POINTERS What is a pointer ?
A pointer is nothing but a variable that contains the address which is a location of the another variable in memory.
Benefits of using the pointer :
A pointer enables us to access a variable that is defined outside the function.
Pointer are more efficient in handling data table.
Pointer reduces the length and complexity of the program.
The use of pointer array to character string results in saving of data storage space in memory.

3. Declaring and initializing pointers :
syntax :
data_type *pt_name ;
This tells the compiler three things about the variable pt_name.
1. * tells the compiler that the variable pt_name is a pointer variable.
2. pt_name needs a memory location.
3. pt_name points to a variable of type data_type.

4. Ex: int quantity=179 ;
int *p;
p=&quantity;
quantity variable
179 value
5000 Address
Note : you can know the address of variable using %u format specification.
You can’t assign an absolute address to a pointer variable directly.
Ex: p = 5000; It is not possible

5. Accessing variables using pointers. void main()
Output:
Value of X is 10
10 is stored at address 4104
4104 is stored at address 4106
10 is stored at address 4108
Now x = 25
Accessing variables using pointers.
void main()
{ int x,y, *ptr;
x=10;
ptr=&x;
y = *ptr;
printf(“Value of x is %d\n\n”,x);
printf(“%d is stored at address %u\n”, x, &x);
printf(“%d is stored at address %u\n”, *&x, &x);
printf(“%d is stored at address %u\n”, *ptr, &x);
printf(“%d is stored at address %u\n”, y, &*ptr);
printf(“%d is stored at address %u\n”,ptr,&ptr);
printf(“%d is stored at address %u\n”, y, &y);
*ptr=25;
printf(“\n Now x = %d\n”,x) }

6. Pointers expression :
void main() {
int a,b,*p1,*p2, x, y, z ;
a =12; b = 4;
p1=&a; p2 = &b;
x = *p1 * *p2 – 6;
printf(“Address of a = %u\n”,p1);
printf(“Address of b=%u\n”,p2);
printf(“\n”);
printf(“a=%d, b=%d\n”,a , b);
printf(“x=%d \n”,x);

7. printf(\n a = %d , b = %d “, a ,b); printf(“z = %d\n”, z); }
*p2 = *p2 + 3 ;
*p1 = *p2 – 5 ;
y = *p1 + *p2;
z= *p1 * *p2 – 6 ;
printf(\n a = %d , b = %d “, a ,b);
printf(“z = %d\n”, z); }
Out put :-
Address of a = 4020
Address of b =4016
a = b=4
x=42 y=9
a=2 b=7 z=8

8. Pointer increments and scale factor :
p1++ will cause the pointer p1 points to the next value of its type.
Ex :- if p1 is an integer pointer with the initial value say 2800,then after the operations p1=p1+1,the value of p1 will be 2802,not 2801
when we increment pointer its value is increased by the length of the data type that it points to. This length is called scale factor.

9. Pointer and Arrays :-
When an array is declared, the compiler allocates a base
Address and sufficient amount of storage to contains all
The elements of the array in memory location.
Base address is the location of the first element of the
array
int x[5] = {1,2,3,4,5}
Element x[0] x[1] x[2] x[3] x[4]
Value
Address
So, base address is,
x = &x[0] = 1000

10. If we declare p as integer pointer, then we can make the pointer p to point to the array x by the following statements.
p = x or p = &x[0]
Now we can access every element of x using p++ to move from one element to another.
p = &x[0] (=1000)
p+1 = &x[1] (=1002)
p+2 = &x[2] (=1004)
p+3 = &x[3] (=1006)
p+4 = &x[4] (=1008)

11. /*Program of sum n array elements*/
#include<stdio.h>
main() {
int x[10],i,*p,n,sum=0;
printf("enter the N:");
scanf("%d",&n);
printf("enter the the data:\n");
for(i=0;i<n;i++)
scanf("%d",&x[i]);
p = x;

12. /*Program of sum n array elements*/
for(i=0;i<n;i++) {
printf("\n%d",*p);
sum=sum + *p;
p++; }
printf("\nsum=%d",sum);
OUTPUT
Enter the N: 5
Enter the data :
Sum=15

13. 2,0 2,3 Array using pointer notation P P + 1 p+2 *(p+2) + 3 *(p+2)
2,0
2,3 P 1
P + 1
p+2 2 3
*(p+2)
*(p+2) + 3
p = pointer to first row
p+i = pointer to ith row
*(p+i) = pointer to first element in the ith row
*(p+i)+j = pointer to jth elements
*(*(p+i)+j)=value stored in the cell(i,j)

14. /* Program of sum of n array elements*/
#include<stdio.h>
void main() {
int x[10][10],i,j,n,sum=0;
clrscr();
printf("enter the N:");
scanf("%d",&n);
printf("enter the the data:\n");
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&x[i][j]);

15. for(i=0;i<n;i++) {
for(j=0;j<n;j++)
printf("\n%d",*(*(x+i)+j));
sum=sum + *(*(x+i)+j); }
printf("\nsum=%d",sum);
getch();

16. Pointer and character string :-
In c, a constant character string always represents a pointer to that string. And therefore you can directly write char *name;
name = “DELHI”
But remember that this type of assignment does not apply on character arrays.
char name[20];

17. char *name[3] = {“New Zealand”,“Australia”,“India”}
Where name to be an array of three pointers to a character, each pointer points to a particular name.
name[0]=“New Zealand”,
name[1]=“Australia”,
name[2]=“India”
This declaration allocates only 28 bytes.
The character arrays with the rows of varying length are called ragged array.

18. /****PROGRAMM FOR STRING PRINT USING POINTER ****/
#include<stdio.h>
void main() {
char name[10],*ptr;
clrscr();
printf("enter the name => ");
scanf("%s",name);
ptr=name;
printf("\n");

19. while(*ptr != '\0') {
printf("%c",*ptr);
ptr++; }
getch();
Note : if we want to directly assign string constant then we have to make name as character pointer.
char *name ; name = “Delhi”