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c++macrospreprocessorvariadic-macros
Mohammed Micro
Mohammed Micro

Reputation: 333

What does ##__VA_ARGS__ mean?

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?

Upvotes: 23

Views: 57436

Answers (2)

waterd
waterd

Reputation: 651

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__)

Upvotes: 2

Trass3r
Trass3r

Reputation: 6287

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

Upvotes: 25

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