CODEWITHSUNDEEP
c++macrosinline
Tuğberk Kaan Duman
Tuğberk Kaan Duman

Reputation: 950

Mixing inline and macro functions

I've decided to try something. I know macros are evil and should be avoided but wanted to see what's going to happen if I do such a thing.

#include <iostream>
using namespace std;
inline void add(int x, int y) { cout << "Inline: " << x + y << endl; }
#define add(x,y) ( cout << "Macro: " << x + y << endl )

int main()
{
    add(3,5);
}

It outputs:

Macro: 8

If I comment out the #define line inline starts working, and the output turns into Inline: 8.

My question is, why compiler decides to use macro function instead of inline. Thank you!

I'm using Linux Mint 18.2, g++ 5.4.0, with no parameters g++ -g t2.cpp -o t2.

Upvotes: 2

Views: 335

Answers (3)

iBug
iBug

Reputation: 37227

A macro comes to effect at the place it is defined, so the inline function is not replaced by the macro, but your call is replaced before compilation. Also note that a macro is not a function, it tells the preprocessor a pattern of replacing texts.

You can run the following command and see the output file (your.i)

g++ -o your.i -E your.cpp
               ^

And you'll find the inline function not affected by the macro :)

Option -E for G++ means "preprocess the source file, but don't compile, don't assemble and don't link".

Upvotes: 2

user7860670
user7860670

Reputation: 37520

There is not such thing as "macro function". Compiler (actually preprocessor) just turns main into

int main()
{
    ( cout << "Macro: " << 3 + 5 << endl );
}

Upvotes: 2

user2100815
user2100815

Reputation:

Macro substitution is performed via the pre-processor before compilation. Thus the compiler never sees add(3,5) - it only sees the macro expansion.

Upvotes: 5

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