Question

The following code prints different results on 32bit and 64bit systems:

#include <stdio.h>
void swapArray(int **a, int **b) 
{ 
    int *temp = *a; 
    *a = *b; 
    *b = temp; 
} 

int main() 
{ 
    int a[2] = {1, 3}; 
    int b[2] = {2, 4}; 
    swapArray(&a, &b); 
    printf("%d\n", a[0]); 
    printf("%d\n", a[1]); 
    return 0; 
}

After compiling it in 32bit system, the output is:

2
3

On 64bit the output is:

2
4

As I understand, the function swapArray just swaps the pointers to the first elements in a and b. So after calling swapArray, a should point to 2 and b should point to 1.
For this reason a[0] should yield 2, and a[1] should reference the next byte in memory after the location of 2, which contains 4.

Can anyone please explain?

Edit:
Thanks to the comments and answers, I now notice that &a and &b are of type int (*)[] and not int **. This obviously makes the code incorrect (and indeed I get a compiler warning). It is intriguing, though, why the compiler (gcc) just gives a warning and not an error.
I am still left with the question what causes different results on different systems, but since the code is incorrect, it is less relevant.

Edit 2:
As for the different results on different systems, I suggest reading AndreyT’s comment.

Answer 1

swapArray(&a, &b); 

&a and &b are not of type int ** but of type int (*)[2]. BTW your compiler is kind enough to accept your program but a compiler has the right to refuse to translate it.