Using #define for a function

Question

After struggling with C++ typedef and define, I've tried myself some code and here it is:

#include <iostream>
#include <string>
#define sub(a,b) a-b
using namespace std;
int main()
{
    int n;
    cin >> n;
    cout <<  n*sub(5*2,3*3)+1;
}

I've find that if n=3, it'll give 22 and the equation is 3*5*2-3*3+1=22;

I can't imagine the mechanism of it, can someone help me please?

`#define sub(a,b) a-b` -> `#define sub(a,b) ((a)-(b))`, but really the best option is to not use a macro. — Retired Ninja, May 13 '22 at 09:49
`#define` is a preprocessor directive and does pure text replacement. It does not evaluate the arguments that are passed. So your equation turns into `n*5*2-3*3` — UnholySheep, May 13 '22 at 09:49

Vlad from Moscow · Answer 1 · 2022-05-13T09:55:29.697

1

Just insert in this expression

n*sub(5*2,3*3)+1

the macro extension 5*2 - 3*3 and you will have

n * 5*2 - 3*3 + 1

You should rewrite the macro at least like

#define sub(a,b) ( (a ) - ( b ) )

And in C++ it is better to use inline functions (possibly with the specifier constexpr) instead pf such a macro.

edited May 13 '22 at 09:55

answered May 13 '22 at 09:49

Vlad from Moscow

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Using #define for a function

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