Conforming way to handle varargs in macro function

Question

I'm trying to understand how to handle varargs in a macro function conformingly (i.e. avoiding gcc's comma swallowing ##__VA_ARGS__). I wrote this:

#define IS_DEFINED(ARG) strlen(#ARG)
#define FOO(a, ...) \
    if(IS_DEFINED(__VA_ARGS__)){\
        printf(#a, __VA_ARGS__)\
    } else {\
        printf(#a)\
    }

But it does not generate a valid program test. FOO(1) expands to

if(strlen("")){ printf("1", ) } else { printf("1") };

which does not compile.

Is there a way to handle a case of vararg macro function with one parameter and vararg?

This is for C++, but the idea is the same https://stackoverflow.com/questions/53986620 . — StoryTeller - Unslander Monica, Jan 13 '19 at 06:45
@StoryTeller It's kindof offtop, buy may I ask you the reason to use `do{}while(0)` construction in the replacement list? Why doesn't a simple block `{ }` do the job? — Some Name, Jan 13 '19 at 06:52
It's for the semi-colon at the end. See here https://stackoverflow.com/questions/154136/why-use-apparently-meaningless-do-while-and-if-else-statements-in-macros — StoryTeller - Unslander Monica, Jan 13 '19 at 06:54
See the discussion in [C `#define` macro for debugging](https://stackoverflow.com/questions/1644868/c-define-macro-for-debug-printing/1644898#1644898) — Jonathan Leffler, Jan 13 '19 at 07:21

score 1 · Answer 1 · answered Feb 13 '23 at 13:23

1

From C23 onwards, you should be able to use __VA_OPT__() for this:


#define FOO(a, ...) \
        printf(#a __VA_OPT__(,) __VA_ARGS__)

answered Feb 13 '23 at 13:23

Toby Speight

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Conforming way to handle varargs in macro function

1 Answers1