#include
void main()
{
int arr[]={1, 2, 3, 4, 5};
int *p=arr;
printf("%d", *p++);
printf("%d", *(p+1));
}
Find the output of the above code?

Question

#include void main() { int arr[]={1, 2, 3, 4, 5}; int *p=arr; printf("%d", *p++); printf("%d", *(p+1)); } Find the output of the above code?

Accepted Answer

Correct answer: B — Answer: 1, 3 Step-by-step explanation: Initial state: p = arr, so p points to arr[0] which has value 1. First printf: *p++ is parsed as *(p++). The dereference uses the original pointer value, so it prints 1. After evaluation, p is incremented and now points to arr[1]. Second printf: *(p+1) takes the current p (which points to arr[1]) and adds 1, so it accesses arr[2], whose value is 3. Final output: 1, 3

Answer

A. 1, 2

Answer

B. 1, 3

Answer

C. 2, 3

Answer

D. 1, 4

#include void main() { int arr[]={1, 2, 3, 4, 5}; int *p=arr; printf("%d",…