How to use __VA_ARGS__ properly

Question

I have define VLog like this:

#define VLog(s, ...)  NSLog(@"%@", [NSString stringWithFormat:(s), ##__VA_ARGS__])

I know VLog(@"hello,%d%@", 1, @"a"); __VA_ARGS__ is replaced with 1, @"a".

Whereas, VLog(@"hello"); __VA_ARGS__ is replaced with what?

and if I define VLog like this:

//delete ##
#define VLog(s, ...)  NSLog(@"%@", [NSString stringWithFormat:(s), ##__VA_ARGS__])`

VLog(@"123"); is pointed out error.

you can use it like printf function. such as VLog (@"test = %@, %@",@"123",@"456"); ##__VA_ARGS__ is array of arguments. (%@, %@) is replace with (s) — larva, Feb 19 '14 at 07:32
See here: http://stackoverflow.com/questions/12053275/what-is-in-c/12055446#12055446 — Richard J. Ross III, Feb 19 '14 at 08:05
@RichardJ.RossIII , the link helps me understand ## nicely, thank you! — simalone, Feb 19 '14 at 08:36

score 0 · Accepted Answer · answered Feb 19 '14 at 08:03

In case of zero variable arguments, __VA_ARGS__ is replaced with nothing. ## is special in that it removes preceding , in that case, so compiler won't complain "expression needed after comma". A macro defined without ##, i.e. [NSString stringWithFormat:(s), __VA_ARGS__] will require additional arguments.

How to use __VA_ARGS__ properly

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