C++ why that code has that output? I would love to ubderstand what is hapening inside

Question

Why this code:

#include <iostream>
#define DIFF(x,y) x-y

int main()
{
    std::cout <<   DIFF(5,3) << std::endl;
    std::cout << 1*DIFF(5,3) << std::endl;
    std::cout << 2*DIFF(5,3) << std::endl;
    std::cout << 3*DIFF(5,3) << std::endl;
    std::cout << 4*DIFF(5,3) << std::endl;
    std::cout << 5*DIFF(5,3) << std::endl;
    std::cout << 6*DIFF(5,3) << std::endl;
    std::cout << 7*DIFF(5,3) << std::endl;
    std::cout << 8*DIFF(5,3) << std::endl;
    std::cout << 9*DIFF(5,3) << std::endl;
    
    return 0;
}

has that output? 2 2 7 12 17 22 27 32 37 42

I cannot undestand what is happening inside and cant find documentation about it

Macros are not functions but a straight text replacement in the source code, so `2*DIFF(5,3)` expands to `2*5-3` — Richard Critten, Mar 01 '23 at 23:19
* has higher precedence than -. Remember that macros are fancy string search and replace before compilation. This is why it is recommended to parenthesize your macro expressions. — Avi Berger, Mar 01 '23 at 23:20

score 0 · Answer 1 · answered Mar 01 '23 at 23:24

DIFF is a c preprocessor definition. The c preprocessor is a text processor that doesn't know anything about c. It basically copy-pastes the definition of DIFF into the code before compilation. That means that

    std::cout << 2*DIFF(5,3) << std::endl;

becomes

    std::cout << 2*5-3 << std::endl;

Which is 10-3 = 7

To solve this you could add parentheses like (x-y) in your define.

Or, since you are using c++, you might want to use a consterxpr function like this:

constexrpr int DIFF(int x,int y){ return x-y;}

which has the advantage of type safety, and namespace scoping etc.

C++ why that code has that output? I would love to ubderstand what is hapening inside

1 Answers1