以下内容是使用VS2015编译的,但在VS2017中失败,并出现以下错误。代码是在执行VS2017中已修复的非标准操作,还是VS2017应该对其进行编译?
#include "stdafx.h"
#include <type_traits>
template <typename E>
constexpr auto ToUnderlying(E e)
{
return static_cast<std::underlying_type_t<E>>(e);
}
template<typename T>
bool constexpr IsFlags(T) { return false; }
template<typename E>
std::enable_if_t<IsFlags(E{}), std::underlying_type_t<E>> operator | (E lhs, E rhs)
{
return ToUnderlying(lhs) | ToUnderlying(rhs);
}
enum class PlantFlags { green = 1, edible = 2, aromatic = 4, frostTolerant = 8, thirsty = 16, growsInSand = 32 };
bool constexpr IsFlags(PlantFlags) { return true; }
int main()
{
auto ored = PlantFlags::green | PlantFlags::frostTolerant;
return 0;
}
错误是:
c:\main.cpp(24): error C2893: Failed to specialize function template 'enable_if<false,_Ty>::type
operator |(E,E)'
with
[
_Ty=underlying_type<_Ty>::type
]
c:\main.cpp(24): note: With the following template arguments:
c:\main.cpp(24): note: 'E=PlantFlags'
c:\main.cpp(24): error C2676: binary '|': 'PlantFlags' does not define this operator or a conversion to a type acceptable to the predefined operator
最佳答案
这绝对是Visual Studio中的错误。它compiles on GCC and Clang。它似乎与作为模板参数评估的constexpr
函数有关。作为临时的解决方法,您可以使模板变量:
template <typename T>
bool constexpr is_flags_v = IsFlags(T{});
template<typename E>
std::enable_if_t<is_flags_v<E>, std::underlying_type_t<E>> operator | (E lhs, E rhs)
{
return ToUnderlying(lhs) | ToUnderlying(rhs);
}
On Godbolt