CODEWITHSUNDEEP
c
Irbis
Irbis

Reputation: 1511

c - pointer arithmetic, unexpected results

Here is an example code:

int* arr = (int*)malloc(2 * sizeof(int));

arr = arr + 1;
*arr = 11;
arr[-1] = 22;

int x = arr[0]; // x = 11
int y = arr[1]; // y = 194759710

After a memory allocation arr pointer is incremented. I have expected to get the following results:

x == 22
y == 11

Could you explain how it works ?

Upvotes: 2

Views: 102

Answers (2)

Mathieu
Mathieu

Reputation: 9689

When you set x and y, you forgot that arr no longer points on allocated memory.

This code do what you want:

#include <stdio.h>
#include <stdlib.h>

int main(void) {

    int* arr = (int*)malloc(2 * sizeof(int));
    int *original = arr;

    arr = arr + 1;

    *arr = 11;
    arr[-1] = 22;

    int x = original [0]; // x = arr[-1]
    int y = original [1]; // y = arr[0]

    printf("%d, %d\n", x, y);
}

Upvotes: 5

Lundin
Lundin

Reputation: 215164

The y assignment is incorrect. You only have 2 allocated items so you cannot first increment the pointer by one, then de-reference that pointer with [1], because that gives a 3rd element which doesn't exist.

Upvotes: 5

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