On 12 June 2014 01:42, Nikos Chantziaras <realnc@xxxxxxxxx> wrote: > On 12/06/14 03:31, Jonathan Wakely wrote: >> >> On 11 June 2014 23:36, Nikos Chantziaras wrote: >>> >>> According to the documentation: >>> >>> >>> https://gcc.gnu.org/onlinedocs/gcc-4.8.2/gcc/Function-Attributes.html#Function-Attributes >>> >>> and more specific, the example about the visibility attribute: >>> >>> void __attribute__ ((visibility ("protected"))) >>> f () { /* Do something. */; } >>> >>> doesn't work for me. I have this C++ function prototype: >>> >>> const SDL_AudioSpec& __attribute__((visibility("default"))) >>> spec(); >>> >>> Compiling this with g++ 4.8.2 gives me: >>> >>> warning: 'visibility' attribute ignored on non-class types >> >> >> I think the attribute binds to the return type here, which is a >> reference (not a class type) so can't be given visibility. >> >> This applies the attribute to the function, not the return type: >> >> const SDL_AudioSpec& spec() __attribute__((visibility("hidden"))); > > > I've found that this also works: > > __attribute__((visibility("hidden"))) const SDL_AudioSpec& spec(); > > The only one that doesn't, is the documented one :-/ Documentation bug? No, the example in the documentation is a function with a void return, and in that case the attribute binds to the function not the return type. Your function does not have a void return type. I don't know why that's different, but both the C and C++ compilers treat the example the same.
