What does ##__VA_ARGS__ mean?

Question

I would like to know what ## does in this macro definition:

#define debug(M, ...) fprintf(stderr,M "\n",##__VA_ARGS __)

I googled for an answer and I came up with the following.

The ## will remove the comma if no variable arguments are given to the macro. So, if the macro is invoked like this

debug("message");

with no quotes, it is expanded to

fprintf(stderr,"message");

not

fprintf(stderr,"message",);

Why is the comma removed?

Trass3r · Accepted Answer · 2020-10-09T11:21:32.543

22

It's a non-portable syntax introduced by gcc to specifically deal with this corner case of passing no arguments at all. Without the ## it would complain about the trailing comma being a syntax error.

https://gcc.gnu.org/onlinedocs/gcc/Variadic-Macros.html

C++20 introduced __VA_OPT__ for this purpose: https://en.cppreference.com/w/cpp/preprocessor/replace

edited Oct 09 '20 at 11:21

answered Oct 19 '18 at 11:49

Trass3r

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So it has nothing to do with the general behavior of ## ,concatenation, right? – Mohammed Micro Oct 19 '18 at 12:00
Yeah it's just abusing that operator. – Trass3r Oct 19 '18 at 13:52
As of C++20 you will be able to do this in a standardized way: `#define debug(M, ...) fprintf(stderr,M "\n" __VA_OPT__(,) __VA_ARGS __)` – Carlo Wood Jun 20 '19 at 18:57
@CarloWood It seems that VS2017 does not support __VA_OPT__(,) – Maverick May 02 '20 at 12:38
@Maverick didn't he just say "As of C++20"? :) – Johann Gerell May 04 '20 at 11:21
1

@JohannGerell Yes, I wanted just to make it clear and bold, that even a pro version of the Visual Studio version 2017 that I own, does not support this macro that I arbitrarily thought in the begging...... – Maverick May 05 '20 at 09:15
For C++20 support you'll almost certainly need 2019. – Trass3r May 05 '20 at 13:07
1

VS2017 won't support C++20 lol. Hopefully VS2020 will, if you're lucky :p. For the documentation see https://en.cppreference.com/w/cpp/preprocessor/replace – Carlo Wood May 05 '20 at 15:23

score 2 · Answer 2 · answered Dec 08 '21 at 11:50

from https://en.cppreference.com/w/cpp/preprocessor/replace

Note: some compilers offer an extension that allows ## to appear after a
comma and before __VA_ARGS__, in which case the ## does nothing when the 
variable arguments are present, but removes the comma when the variable
arguments are not present: this makes it possible to define macros such as
fprintf (stderr, format, ##__VA_ARGS__)

What does ##__VA_ARGS__ mean?

2 Answers2

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