void_t在参数列表工作,但不作为返回类型 [英] void_t in parameter list works but not as return type
问题描述
有关于使用别名的cppreference的示例。此示例失败,因为 int
没有成员 foo
:
There's an example on cppreference about the using alias. This example fails because int
has no member foo
:
template<typename...> using void_t = void;
template<typename T> void_t<typename T::foo> f();
f<int>(); // error, int does not have a nested type foo
这很清楚, void_t
部分在参数列表中意外编译:
This is clear, but when I tried putting the void_t
part in the parameter list it unexpectedly compiled:
template<typename...> using void_t = void;
template<typename T> void f(void_t<typename T::foo>);
f<int>();
它在clang上编译,但不在gcc中。这是一个错误吗?
It compiles on clang but not in gcc. Is this a bug?
推荐答案
template<class...>struct voider{using type=void;};
template<class...Ts>using void_t=typename voider<Ts...>::type;
在C ++ 11标准中有一个含混不清之处,即不使用模板参数来模板使用
there is an ambiguity in the C++11 standard about whether non-used template parameters to template using aliases that are invalid types/expressions are a substitution failure or not.
gcc和clang对该子句的解释不同,这是我认为你看到的。上面的 void_t
应该使它在gcc和clang中都是一样的。
gcc and clang interpreted the clause differently, which is what I think you are seeing. The above void_t
should make it work the same in both gcc and clang.
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