CODEWITHSUNDEEP
carrayspointers
Ishtiaq
Ishtiaq

Reputation: 71

Can someone explain what's happening with this code?

#include<stdio.h>

int main()
{
    int a[] = {10, 20, 30, 40, 50};
    int *b = a - 1;

    printf("%d \n",*(a+2));
}

I know that it prints 30 which is same as a[2], but how? What does a - 1 do to the the array a[]?

Upvotes: 1

Views: 69

Answers (1)

M.M
M.M

Reputation: 141576

a - 1 does not change a, in the same way that 3 + 2 does not change 3.

This code causes undefined behaviour because a - 1 tries to form a pointer outside of the bounds of a. But in practice it is likely that the b line will just be ignored, so your code will behave the same as:

int a[] = {10, 20, 30, 40, 50};
printf("%d \n",*(a+2));

which of course prints 30.

Upvotes: 4

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