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Toot

Written by stefanct on 2024-08-10 at 16:14

Playing around with #C23 stuff and got surprised by this not working…

void func(void);

typedef typeof(func) func_type;

Since a function name is neither an expression nor a type name, it's not supported as operand to typeof (not sure that's all), fair enough but why? @AaronBallman @thephd

I'd have used it to have the prototypes of a library functions in a header and use the typedef within the "private" .c only. This would avoid the alternative of duplicating the prototype within the typedef.

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Written by stefanct on 2024-08-10 at 16:27

@AaronBallman @thephd oh it's even worse, typedef doesnt work with typeof at all AFAICT. or i am stupid. >< *opening the actual standard

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Written by Björkus "No time_t to Die" Dorkus on 2024-08-10 at 19:50

@stefanct @AaronBallman This works across all distributions. I think you need to check your flags.

https://godbolt.org/z/z8YTac4x6

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Written by stefanct on 2024-08-11 at 14:23

@thephd @AaronBallman it was actually simply too old compilers. since typeof was an extension for so long, I got completely thrown off the rails. gcc supports it as a non-extension only since version 13.1.0*. thank you and sorry for bothering you with nonsense :(

• https://github.com/gcc-mirror/gcc/commit/fa258f6894801aef6785f0327594dc803da63fbd

(and my initial assumption that a function name isn't an expression was completely bogus too of course - identifiers are primary-expressions).

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