Question:
What will be the output of the following C code?
# include <stdio.h>
int main()
{ int arr[5] = {1, 2, 3, 4, 5};
int *ptr = arr;
*(ptr + 2) = *(ptr + 4) + *(ptr + 1);
printf("%d\n", arr[2]);
return 0; }
What will be the output of the following C code?
# include <stdio.h>
int main()
{ int arr[5] = {1, 2, 3, 4, 5};
int *ptr = arr;
*(ptr + 2) = *(ptr + 4) + *(ptr + 1);
printf("%d\n", arr[2]);
return 0; }
# include <stdio.h>
int main()
{ int arr[5] = {1, 2, 3, 4, 5};
int *ptr = arr;
*(ptr + 2) = *(ptr + 4) + *(ptr + 1);
printf("%d\n", arr[2]);
return 0; }
Show Hint
Rewrite each *(ptr + k) as arr[k] and compute.
Updated On: Jul 2, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Set up the array. arr holds {1, 2, 3, 4, 5} at indices 0 to 4, and ptr points to arr[0].
Step 2: Read the right side of the assignment. *(ptr + 4) is arr[4] which is 5. *(ptr + 1) is arr[1] which is 2.
Step 3: Add them. \(5 + 2 = 7\).
Step 4: The left side *(ptr + 2) is arr[2]. So arr[2] is now set to 7.
Step 5: printf prints arr[2], which is 7. The answer is 7 (option C).
Step 2: Read the right side of the assignment. *(ptr + 4) is arr[4] which is 5. *(ptr + 1) is arr[1] which is 2.
Step 3: Add them. \(5 + 2 = 7\).
Step 4: The left side *(ptr + 2) is arr[2]. So arr[2] is now set to 7.
Step 5: printf prints arr[2], which is 7. The answer is 7 (option C).
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