Order of local variable allocation on the stack

Question

Take a look at these two functions: ``` void function1() { int x; int y; int z; int *ret; } void function2() { char buffer1[4]; char buffer2[4]; char buffer3[4]; int *ret; } ``` If I break at `function1()` in `gdb`, and print the addresses of the variables, I get this: ``` (gdb) p &x $1 = (int *) 0xbffff380 (gdb) p &y $2 = (int *) 0xbffff384 (gdb) p &z $3 = (int *) 0xbffff388 (gdb) p &ret $4 = (int **) 0xbffff38c ``` If I do the same thing at `function2()`, I get this: ``` (gdb) p &buffer1 $1 = (char (*)[4]) 0xbffff388 (gdb) p &buffer2 $2 = (char (*)[4]) 0xbffff384 (gdb) p &buffer3 $3 = (char (*)[4]) 0xbffff380 (gdb) p &ret $4 = (int **) 0xbffff38c ``` You'll notice that in both functions, `ret` is stored closest to the top of the stack. In `function1()`, it is followed by `z`, `y`, and finally `x`. In `function2()`, `ret` is followed by `buffer1`, then `buffer2` and `buffer3`. Why is the storage order changed? We're using the same amount of memory in both cases (4 byte `int`s vs 4 by…

Accepted Answer

I've no idea why GCC organizes its stack the way it does (though I guess you could crack open its source or this paper and find out), but I can tell you how to guarantee the order of specific stack variables if for some reason you need to. Simply put them in a struct:

```
void function1() {
    struct {
        int x;
        int y;
        int z;
        int *ret;
    } locals;
}
```

If my memory serves me correctly, spec guarantees that `&ret > &z > &y > &x`. I left my K&R at work so I can't quote chapter and verse though.

Order of local variable allocation on the stack

Solution

Problem