Method 1
#include <stdio.h>
#include <stdlib.h>
int main()
{
int *ptr,sum=0,n,i;
int a[10];
ptr=a;
printf(“enter the number of elements\n”);
scanf(“%d”,&n);
printf(“enter the array elements\n”);
for(i=0;i<n;i++)
scanf(“%d”,&a[i]);
printf(“MEMORY LOCATION OF THE FIRST VALUE %u\n”,&ptr[0]);
i=0;
while(i<n)
{
sum=sum+(*ptr);
i++;
ptr++;
}
printf(“sum is %d\n”,sum);
printf(“FINAL MEMORY LOCATION AFTER ACCESSING %d INTEGERS IS ptr=%u”,n,ptr);
return 0;
}
Output
enter the number of elements
5
enter the array elements
1
2
3
4
5
MEMORY LOCATION OF THE FIRST VALUE 2752216
sum is 15
FINAL MEMORY LOCATION AFTER ACCESSING IN 5 INTEGERS ptr=2752236
Method 2
#include <stdio.h>
#include <malloc.h>
void main()
{
int i, n, sum = 0;
int *a;
printf("Enter the size of array A \n");
scanf("%d", &n);
a = (int *) malloc(n * sizeof(int));
printf("Enter Elements of First List \n");
for (i = 0; i < n; i++)
{
scanf("%d", a + i);
}
for (i = 0; i < n; i++)
{
sum = sum + *(a + i);
}
printf("Sum of all elements in array = %d\n", sum)
}
Output
Enter the size of array A
5
Enter Elements of First List
4
9
10
56
100
Sum of all elements in array = 179