How C Preprocessor expand the macro with ##?

Question

English is not my native language; thus I show the code to depict.

#define concat_temp(x, y) x##y
#define concat(x, y) concat_temp(x, y)
#define BAR 2

int main() {
    concat_temp(FOO, BAR)
    concat(FOO, BAR)
}

When I executed clang -E, the macros were expanded to

int main() {
    FOOBAR
    FOO2
}

Who can explain to me why concat_temp couldn't expand the bar to 2 but concat did the trick?

Nice! Never know about such things. But also good, that I did not know it because I avoid macros at all :-) — Klaus, Aug 01 '22 at 12:07
Funfact, `concat_temp(FOO, (BAR))` could still yield `FOO2` i think. Could be wrong, and compiler-dependant however. — Shark, Aug 01 '22 at 12:34
@unlsycn alright, point taken, i didn't mess with this macro stuff in like a decade. guess compilers evolved since then :D — Shark, Aug 01 '22 at 13:11

score 5 · Answer 1 · answered Aug 01 '22 at 12:10

5

According to CppReference,

A ## operator between any two successive identifiers in the replacement-list runs parameter replacement on the two identifiers (which are not macro-expanded first) and then concatenates the result.

The ## operator operates on the identifiers themselves, so another pass of macro expansion is required if you want to "make string literal" or concatenate the values of #define-d macros.

answered Aug 01 '22 at 12:10

iBug

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Avoided my dupe hammer by 9 seconds ;D – Aykhan Hagverdili Aug 01 '22 at 12:11
1

@AyxanHaqverdili In fact, by [4 more hours](https://meta.stackexchange.com/a/316139/350567) available. – iBug Aug 01 '22 at 12:12
Wow, [it did work](https://stackoverflow.com/a/73193542/10147399). I am going to abuse this power... – Aykhan Hagverdili Aug 01 '22 at 12:19

How C Preprocessor expand the macro with ##?

1 Answers1