Why won't my variadic macro accept no arguments correctly?

Question

Overloading Macro on Number of Arguments

https://codecraft.co/2014/11/25/variadic-macros-tricks/

I've been looking at the two links above, trying to get the following code to work:

#define _GET_NUMBER(_0, _1, _2, _3, _4, _5, NAME, ...) NAME
#define OUTPUT_ARGS_COUNT(...) _GET_NUMBER(_0, ##__VA_ARGS__, 5, 4, 3, 2, 1, 0)

...

cout << OUTPUT_ARGS_COUNT("HelloWorld", 1.2) << endl;
cout << OUTPUT_ARGS_COUNT("HelloWorld") << endl;
cout << OUTPUT_ARGS_COUNT() << endl;

This compiles, runs, and gives the following output:

2
1
1

I can not for the life of me figure out why the call OUTPUT_ARGS_COUNT() is giving me 1 instead of 0. I have an ok understanding of the code I'm trying to use, but it's a tad greek to me still so I guess it's possible that I'm not applying something correctly despite the fact I literally copied and pasted the example code from the link on stack overflow.

I'm compiling using g++ 5.4.0 20160609.

Any ideas or additional resources you can point me to would be greatly appreciated.

You can see at gcc.gnu.org/onlinedocs/cpp/Variadic-Macros.html: "The above explanation is ambiguous about the case where the only macro parameter is a variable arguments parameter, as it is meaningless to try to distinguish whether no argument at all is an empty argument or a missing argument. CPP retains the comma when conforming to a specific C standard. Otherwise the comma is dropped as an extension to the standard.". — Jarod42
Check this stackoverflow.com/questions/2308243/…, the solution there also fails on zero arguments — Passer By

Jarod42 Jarod42 · Accepted Answer · 2017-10-17T15:53:14

You can see at http://gcc.gnu.org/onlinedocs/cpp/Variadic-Macros.html:

Second, the ‘##’ token paste operator has a special meaning when placed between a comma and a variable argument. If you write

#define eprintf(format, ...) fprintf (stderr, format, ##__VA_ARGS__)

and the variable argument is left out when the eprintf macro is used, then the comma before the ‘##’ will be deleted. This does not happen if you pass an empty argument, nor does it happen if the token preceding ‘##’ is anything other than a comma.

eprintf ("success!\n")
 → fprintf(stderr, "success!\n");

The above explanation is ambiguous about the case where the only macro parameter is a variable arguments parameter, as it is meaningless to try to distinguish whether no argument at all is an empty argument or a missing argument. CPP retains the comma when conforming to a specific C standard. Otherwise the comma is dropped as an extension to the standard.

So, (unless appropriate extension used) OUTPUT_ARGS_COUNT() is counted as 1 empty argument (comma kept with ##__VA_ARGS__).

Why won't my variadic macro accept no arguments correctly?

2 Answers