请考虑以下代码：
struct D
{
template <class T> D (T);
};
int operator~(const D &);
template <typename T> T &make ();
template <typename Rhs> struct H
{
static const int value = sizeof ~make<Rhs>();
};
enum class E;
int main () { return H<E>::value; }
这是有效的C 11吗？
Clang接受了它. Gcc给出了一个错误：
% gcc -std=c++11 b.ii b.ii: In instantiation of ‘const int H<E>::value’: b.ii:16:28: required from here b.ii:11:35: error: no match for ‘operator~’ (operand type is ‘E’) static const int value = sizeof ~make<Rhs>();
代码从gcc错误报告中减少：
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=60852
这是未减少的测试用例：
#include <boost/type_traits.hpp>
#include <iostream>
enum class E {};
int main()
{ std::cout << boost::has_complement<E>() << std::endl; }
解决方法
GCC就在这里,范围枚举,aka枚举类不会隐式转换为int或其他任何东西.通过扩展,它们没有内置运算符〜,因此您需要显式转换：
#include <iostream>
#include <type_traits>
enum class E { val };
int main () {
std::cout << ~std::underlying_type<E>::type(E::val);
}
通过删除struct D和全局运算符〜,clang给出了错误.这是一个明显的错误,因为运算符〜(D)首先不是候选人：
main.cpp:17:35: error: invalid argument type 'E' to unary expression static const int value = sizeof ~make<Rhs>();
关于收集过载的规则§13.3.1.2：
For a unary operator @ with an operand of a type whose cv-unqualified
version is T1,[…],three sets of candidate functions,designated
member candidates,nonmember
candidates and built-in candidates,are constructed as follows:
结束：
However,if no operand has a class type,only those non-member
functions in the lookup set that have a first parameter of type T1 or
reference to (possibly cv-qualified) T1”,when T1 is an enumeration
type,are candidate functions.
总而言之,因为E是非类型的,所以只考虑采用E或E的引用的自由函数运算符.
