你去:
#include <type_traits>
#include <iostream>
#include <vector>
template<typename T>
struct is_const_mem_fn {
private:
template<typename U>
struct Tester {
static_assert( // will always fail
std::is_member_function_pointer<U>::value,
"Use member function pointers only!");
// if you want to report false for types other than
// member function pointers you can just derive from
// std::false_type instead of asserting
};
template<typename R, typename U, typename...Args>
struct Tester<R (U::*)(Args...)> : std::false_type {};
template<typename R, typename U, typename...Args>
struct Tester<R (U::*)(Args...) const> : std::true_type {};
public:
static const bool value =
Tester<typename std::remove_cv<T>::type>::value;
};
struct A {
int member();
int member() const;
};
typedef int (A::*PtrToMember)();
typedef int (A::*PtrToConstMember)() const;
int main()
{
std::cout
<< is_const_mem_fn<PtrToMember>::value
<< is_const_mem_fn<const PtrToMember>::value
<< is_const_mem_fn<PtrToConstMember>::value
<< is_const_mem_fn<const volatile PtrToConstMember>::value
<< is_const_mem_fn<decltype(&std::vector<int>::size)>::value;
}
输出:00111
编辑:我忘记在原始答案中说明一个极端情况。
上面的 trait 会扼杀一个假设的成员函数,如下所示:
struct A {
int member(int, ...) const;
};
因为没有Tester
可以为此类签名生成的有效专业化。要修复它,请添加以下专业化:
template<typename R, typename U, typename...Args>
struct Tester<R (U::*)(Args..., ...)> : std::false_type {};
template<typename R, typename U, typename...Args>
struct Tester<R (U::*)(Args..., ...) const> : std::true_type {};