Variadic macro with no arguments for its variadic parameter

Question

Is it legal to invoke a variadic macro M with no arguments for its variadic parameter?

The relevant standard quote is [cpp.replace]/4:

If the identifier-list in the macro definition does not end with an ellipsis, the number of arguments (including those arguments consisting of no preprocessing tokens) in an invocation of a function-like macro shall equal the number of parameters in the macro definition. Otherwise, there shall be more arguments in the invocation than there are parameters in the macro definition (excluding the ...). There shall exist a ) preprocessing token that terminates the invocation.

For the case with no non-variadic parameters, an invocation in the form M() should be legal as the invocation has one argument (consisting of no preprocessing tokens). So there is one more argument than non-variadic parameters.

For the case with one non-variadic parameter, should there be a trailing , as in M(1,) to introduce an argument consisting of no preprocessing tokens for the variadic parameter? Otherwise, the number of arguments would be equal to the number of non-variadic parameters. i.e.,

#define variadic(x,...) #__VA_ARGS__

variadic(1,) // 2 arguments: ok
variadic(1) // 1 argument: wrong?

Both Clang and GCC, however, accept the following test case:

#include <iostream>

#define variadic(x,...) #__VA_ARGS__

int main()
{
    std::cout << "'" variadic(1) "'" << std::endl;
}

and produce as output:

''

Is this non-standard behavior?

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2017/p0306r4.html has been merged in C++2a's draft, changing `there shall be more arguments in the invocation` to `there shall be at least as many arguments`, thus making the above code legitimate. — peppe, Dec 20 '18 at 20:03

score 8 · Accepted Answer · answered May 27 '13 at 08:30

8

Otherwise, there shall be more arguments in the invocation than there are parameters in the macro definition (excluding the ...).

This very extract from the standard shows that your code should not be valid: you have one parameter plus the ellipsis. If we follow the portion of the standard above, you should have at least two arguments. When you write varidadic(1), you just provide one argument. Your code is not valid.

By the way, clang produces a warning:

main.cpp:7:32: warning: must specify at least one argument for '...' parameter of variadic macro [-Wgnu]
    std::cout << "'" variadic(1) "'" << std::endl;

And GCC also produces a warning:

main.cpp:7:32: warning: ISO C99 requires rest arguments to be used [enabled by default]
     std::cout << "'" variadic(1) "'" << std::endl;

Since that may be a bother to the programmer, and since the programmer's intent is easy to guess, they both consider variadic(1) is equivalent to variadic(1,).

answered May 27 '13 at 08:30

Morwenn

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1

What should one do to eliminate these warnings? – rr- Jun 14 '15 at 15:17
@rr- Fix the problem: either consistently pass two arguments or make the macro take "at least 0" arguments instead of "at least one". – Morwenn Jun 15 '15 at 14:46
5

Is changing `#define variadic(x,...) #__VA_ARGS__` → `#define variadic(...) #__VA_ARGS__` sufficient? (example taken from the OP's post) – rr- Jun 15 '15 at 14:47
1

@rr- It is, unless `x` is intended to be used separately from the rest of the arguments. – Morwenn Jun 15 '15 at 15:04

Variadic macro with no arguments for its variadic parameter

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