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The name of an array refers to the address of the first element of the array, i.e. an expression of array type decomposes to pointer type
int a[3] = {10,20,30};Here,
a[0] = 10
a[1] = 20
a[2] =30the same can be written as
*(a+0) = 10
*(a+1) = 20
*(a+2) = 30printf(“Address of array = %u”, a)The above statement prints the address of first element of array ‘a’
int *ptr = a; // Assigning address of a to ptr
*ptr = 10
*(ptr + 1) = 20In general, *(ptr + i) gives the value of element in i th location.
#include<stdio.h>
int main()
{
int arr[3] = {10,15,20};
printf("first element is at %p\n",arr);
printf("second element is at %p\n",arr+1);
printf("third element is at %p\n",arr+2);
return 0;
}Output
first element is at 000000000062FE10
second element is at 000000000062FE14
third element is at 000000000062FE18Program: C program to get the array element using pointer and print it using pointer
#include<stdio.h>
int main()
{
int a[100];
int n,i;
int *ptr;
printf("Enter size of array\n");
scanf("%d",&n);
ptr = a; // Initializing pointer with starting address of array a
// loop to read array elements
printf("Read Array Elements\n");
for(i = 0;i<n;i++)
{
scanf("%d",ptr);
ptr++;
}
//loop to print array elements
ptr = a;
printf("Array Elements are\n");
for(i=0;i<n;i++)
{
printf("%d\t",*ptr);
ptr++;
}
return 0;
}Output
Enter size of array
5
Read Array Elements
3 4 2 1 6
Array Elements are
3 4 2 1 6Program: C program to read array elements using pointer and print it in reverse order
#include<stdio.h>
int main()
{
int a[100];
int n,i;
int *ptr;
printf("Enter size of array\n");
scanf("%d",&n);
ptr = a; // Initializing pointer with starting address of array a
// loop to read array elements
printf("Read Array Elements\n");
for(i = 0;i<n;i++)
{
scanf("%d",ptr);
ptr++;
}
ptr = &a[n-1];
//loop to print array elements
printf("Array Elements are\n");
for(i=n-1;i>=0;i--)
{
printf("%d\t",*ptr);
ptr--;
}
return 0;
}Output
Enter size of array
6
Read Array Elements
1 4 2 3 9 5
Array Elements are
5 9 3 2 4 1The expression of the form E1[E2][E3] is automatically converted into an equivalent expression of the form *(*(E1+E2)+E3)
#include<Stdio.h>
int main()
{
int a[2][3] = {2,1,3,2,3,4};
printf("Using subscript operator\n");
printf("%d\t%d\t%d\n",a[0][0], a[0][1], a[0][2]);
printf("%d\t%d\t%d\n",a[1][0], a[1][1], a[1][2]);
printf("Using pointer \n");
printf("%d\t%d\t%d\n",*(*(a+0)+0),*(*(a+0)+1), *(*(a+0)+2) );
printf("%d\t%d\t%d\n",*(*(a+1)+0),*(*(a+1)+1), *(*(a+1)+2));
printf("Using mixed form \n");
printf("%d\t%d\t%d\n",*(a[0]+0),*(a[0]+1), *(a[0]+2) );
printf("%d\t%d\t%d\n",*(a[1]+0),*(a[1]+1), *(a[1]+2) );
return 0;
}Output
Using subscript operator
2 1 3
2 3 4
Using pointer
2 1 3
2 3 4
Using mixed form
2 1 3
2 3 4- The expression of the form E1[E2][E3] is automatically converted into an equivalent expression of the form *(*(E1+E2)+E3)
- The expression *(a+i) is equivalent to a[i].
- Hence the expression *(*(a+i)+j) is equivalent to *(a[i] +j) which is further equivalent to a[i][j]
- For a two dimensional array,
- When no subscript is used, the expression array name refers to the starting address of the first element(i.e. first row of the array
- When one subscript is used , the expression a[0] and a[1] represents the starting address of the array.
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