Argument counting in macro

问题:

I'm trying to understand the argument counting in C preprocessing macro and the idea in this answer. We have the following macro (I changed the number of arguments for simplicity):

#define HAS_ARGS(...) HAS_ARGS_(__VA_ARGS__, 1, 1, 0,)
#define HAS_ARGS_(a, b, c, N, ...) N

As far as I understand the purpose of this macro is to check if the given varargs empty. So on empty varargs the macro invokation is replaced with 0 which seems fine. But with a single argument it also turns into 0 which I seems strange.

HAS_ARGS(); //0
HAS_ARGS(123); //also 0
HAS_ARGS(1, 2); //1

LIVE DEMO

I think I understand the reason. In case of empty varargs a is replaced with empty preprocessing token, in case of a single argument vararg a is replaced with the argument yielding the same result.

Is there a way to get 0 returned in case varargs are empty, 1 in case argument number is from 1 to the defined in HAS_ARGS_ macro invokation without using comma-swallowing or other non-conforming tricks. I mean

SOME_MACRO_F() //0
SOME_MACRO_F(234) //1
SOME_MACRO_F(123, 132) //1
//etc

回答1:

You cannot pass zero arguments to HAS_ARGS(...). ISO C (and C++, at least for the next two years) requires that an ellipsis corresponds to at least one additional argument after the last named one.

If there are no named ones, then the macro needs to be passed at least one argument. In the case of HAS_ARGS() the extra argument is simply an empty token sequence. Zero arguments is simply not possible.

This is exactly the use case in the answer. The target macro expects at least one argument. So we can use a wrapper accepting only an ellipsis for "overload resolution". A better name probably would have been HAS_MORE_THAN_1_ARGS. Because that's what the predicate is meant to tell you. Alas, I favored brevity on that answer.

回答2:

It seems difficult to compute that at compile-time, but you can do it at run-time by stringifying the arguments and testing if the string is empty.

Tested with gcc:

#include <stdio.h>
#define HAS_ARGS(...) (#__VA_ARGS__[0] != '\0')

int main()
{
   printf("%d %d %d %d\n",HAS_ARGS(),HAS_ARGS(10),HAS_ARGS(20,"foo"),HAS_ARGS(10,20));
    return 0;
}

this prints:

0 1 1 1

behind the scenes, here's what the pre-processor outputs:

int main()
{
   printf("%d %d %d %d\n",(("")[0] != '\0'),(("10")[0] != '\0'),(("20,\"foo\"")[
0] != '\0'),(("10,20")[0] != '\0'));
    return 0;
}