[Animats]

Yes. This only works for arrays on the stack, at best. It assumes that arrays are placed on the stack in the order of declaration, which is not a requirement of the C standard and may differ between compilers.

Unless you're writing a buffer overflow exploit, in which case you need to know exactly what's on the stack and where, this isn't a good way to program.

Update: misread the article; thought he was differencing with the beginning of the next array.

[dllthomas]

I don't see how the code assumes anything about the placement of the array. Indeed, it works just fine for static arrays:

    $ cat test.c
    #include <stdio.h>
    
    int arr[5];
    
    int main(int argc, char *argv[]) {
    	printf("%lu, %ld\n", sizeof(arr) / sizeof(*arr), (&arr)[1] - arr);
    }
    
    $ gcc test.c && ./a.out
    5, 5

Not saying it's "a good way to program" - it's needlessly obfuscated compared to the standard sizeof alternative. But it doesn't rely on anything tricky.

[ycmbntrthrwaway]

> It assumes that arrays are placed on the stack in the order of declaration

I am not sure it is the case here. The code uses only one array, how can it assume the order of arrays?