C++ macro: absolute value?

Question

dumb question. Just started using C++ macros, but can't understand what's wrong here.

#include <iostream>
using namespace std;

#define ABS(a) ((a < 0) ? (-a) : (a))

int main() {
    uint16_t a = 0xf0, b = 0xf8;
    cout << a << " " << b << " " << a - b <<" " << ABS(a - b) << endl;
    int c = a, d = b;
    cout << c << " " << d << " " << c - d << " " << ABS(c - d) << endl;
}

The output is following:

240 248 -8 -488
240 248 -8 -488

Tried with signed and unsigned, but the macro result is always the same.

Welcome to Stack Overflow. You have discovered one of the biggest reasons not to use macros. You know what `240 - 248` is; what happens when you put '-' in front of it? — Beta, Jul 03 '21 at 16:24
Indeed, this should not be a macro to begin with. Why not making it a normal function? — spectras, Jul 03 '21 at 16:42
Macros are evil. What is preventing somebody to use `ABS('&')` or ABS(std::vector)? Perfer to use library functions or inline functions. — Thomas Matthews, Jul 03 '21 at 17:03

score 5 · Answer 1 · answered Jul 03 '21 at 16:25

Let's take this macro:

#define ABS(a) ((a < 0) ? (-a) : (a))

And let's expand just this usage of this macro:

ABS(a - b)

If you go ahead and substitute verbatim (because that's how macros work) a-b everywhere a occurs in the macro, this is what you get:

((a-b < 0) ? (-a-b) : (a-b))

You can pretty much use your text editor's search/replace function to do this. The mistake is now glaringly obvious.

Your macro should be:

#define ABS(a) (((a) < 0) ? -(a) : (a))

(I threw in an extra set of parenthesis, just to avoid further trouble down the road).

C++ macro: absolute value?

1 Answers1