Given the `_fxsave()` function returning `void`, it is invalid C but valid C++:
# WG14 N1256 (C99) / N1570 (C11) 6.8.6.4 The return statement Constraints 1 A return statement with an expression shall not appear in a function whose return type is void. ... # WG21 N1804 (C++03) 6.6.3 The return statement [stmt.return] 3 A return statement with an expression of type “cv void” can be used only in functions with a return type of cv void; the expression is evaluated just before the function returns to its caller. # WG21 N4582 (C++1z) 6.6.3 The return statement [stmt.return] 2 ... A return statement with an operand of type void shall be used only in a function whose return type is cv void. ... ------------------ Best regards, lh_mouse 2016-08-19 ------------------------------------------------------------- 发件人：David Wohlferd <d...@limegreensocks.com> 发送日期：2016-08-19 11:51 收件人：gcc@gcc.gnu.org 抄送： 主题：fxsrintrin.h According to the docs (https://gcc.gnu.org/onlinedocs/gcc/x86-Built-in-Functions.html), __builtin_ia32_fxsave() has return type 'void.' Given that, does this code (from gcc/config/i386/fxsrintrin.h) make sense? _fxsave (void *__P) { return __builtin_ia32_fxsave (__P); } Returning a void? Is that a thing? Similar question for _fxrstor, _fxsave64, and _fxrstor64. And again in xsaveintrin.h for _xsave, _xrstor, _xsave64 and _xrstor64? dw
