Question

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Confused about C macro expansion and integer arithmetic
A riddle (in C)

The expected output of the following C program is to print the elements in the array. But when actually run, it doesn’t do so.

#include<stdio.h>

  #define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
  int array[] = {23,34,12,17,204,99,16};
  int main() 
  {
      int d;
      for(d=-1;d <= (TOTAL_ELEMENTS-2);d++)
          printf("%d\n",array[d+1]);
      return 0;
  }

Answer 1

Because sizeof gives you an unsigned value, which you probably would have noticed had you turned up the warning level, such as using -Wall -Wextra with gcc ^(a):

xyzzy.c: In function 'main':
xyzzy.c:8: warning: comparison between signed and unsigned

If you force it to signed, it works fine:

#define TOTAL_ELEMENTS (int)((sizeof(array) / sizeof(array[0])))

---

What happens in detail can be gleaned from the ISO standard. In comparisons between different types, promotions are performed to make the types compatible. The compatible type chosen depends on several factors such as sign compatibility, precision and rank but, in this case, it was deemed that the unsigned type size_t was the compatible type so d was upgraded to that type.

Unfortunately, casting -1 to an unsigned type (at least for two’s complement which is almost certainly what you’re using) results in a rather large positive number.

One that’s certainly larger the the 5 you get from (TOTAL_ELEMENTS-2). In other words, your for statement effectively becomes:

for (d = some big honking number way greater than five;
     d <= 5;
     d++
) {
    // fat chance of getting in here !!
}

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^(a) This requirement to use extra remains a point of contention between the gcc developers and myself. They’re obviously using some new definition of the word “all” of which I was previously unaware (with apologies to Douglas Adams).