C++ #define directive

Question

There is the next code:

#include <iostream>
#define F(x) (2*x*x)
using namespace std;

int main()
{
    int a = 1, b = 2;
    int res = F(a + b);
}

The res value is 6, but why is it?

Macros are simple text replacement. Your macro gets expanded to `(2*a+b*a+b)` — Yksisarvinen, Nov 19 '21 at 17:57
[Macros are evil!](https://stackoverflow.com/q/14041453/10077) — Fred Larson, Nov 19 '21 at 18:03

score 3 · Answer 1 · answered Nov 19 '21 at 18:00

3

Macros are used to replace the text. This is what ht code is doing:

2*x*x

replacing x with (a+b) 2*a+b*a+b

a = 1 b = 2

Answer is 6

answered Nov 19 '21 at 18:00

Abhishek Dutt

1,308
7
14
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score 0 · Answer 2 · answered Nov 19 '21 at 18:00

The #define directive simply replace the parameter in the macro. So in your code:

F(a + b) turns into 2*a + b*a + b which is 2*1+2*1+2 that is equal to 6.

I suggest to use inline functions for this matter, but if you really want to use a macro (which is highly unrecommended for cases like these for many reasons) surround each parameter with brackets.

Example: #define F(x) (2*(x)*(x))

C++ #define directive

2 Answers2