Using preprocessor macros to compose another macro call

Question

Suppose I have a C++ preprocessor macro defined as follows:

#define X(s) std::cout << #s

if I use it directly:

int main() {
    X( hello );
}

It works as expected and "hello" is printed on the console.

If I define another macro that calls it:

#define Y X( hello )
#define X(s) std::cout << #s

int main() {
    Y;
}

It still works.

However if I try to compose the call to X from two or more different macros, I get a whole bunch of errors:

#define A X(
#define B hello
#define C )

#define X(s) std::cout << #s << '\n'


int main()
{
    A B C;
}

See output at: http://cpp.sh/5ws5k

Why can't I compose a macro call from two or more macro expansions, doesn't preprocessor expand them recursively?

You should read a documentation describing the C++ preprocessor, perhaps the C++11 or C99 standard. It does not work like you are dreaming. — Basile Starynkevitch
– Basile Starynkevitch, Commented Jan 6, 2016 at 16:02

R Sahu · Accepted Answer · 2016-01-06 16:33:50Z

9

Why can't I compose a macro call from two or more macro expansions, doesn't preprocessor expand them recursively?

You can compose macros. The pre-processor does expand macros recursively.

However, it doesn't expand the macros width first. It expands them depth first.

You are running into a problem because you want the pre-processor to expand the macros width first.

You can read more about recursive macro expansion in 16.3.4 Rescanning and further replacement of the C++11 standard.

edited Jan 6, 2016 at 16:33

answered Jan 6, 2016 at 16:05

R Sahu

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2 Comments

imreal Over a year ago

Relevant C++ Standard quote: 16.3.4 Rescanning and further replacement

Potatoswatter Over a year ago

As for "depth" and "width," the preprocessor does both. Moving the open paren from A to B, and given #define I(Q) Q, and using I(A B C), it works.

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