Casting from member function pointer to another type and back, strict-aliasing issue?
铸造成员函数指针到另一个类型和背部，严格的混叠问题？
Determines whether a given type is a pointer to member object type.
判定一个给定的类型是否为成员对象指针类型。
If the base member took different arguments than the derived-class member, there would be no way to call the derived function from a reference or pointer to the base type.
如果基类成员与派生类成员接受的实参不同，就没有办法通过基类类型的引用或指针调用派生类函数。
This error can be caused by a missing type specifier in the declaration of a pointer to a member.
造成这项错误的原因可能是：在成员指标的宣告中遗漏型别规范。
Only arithmetic, enumeration, pointer, pointer to member, and class type arguments are allowed.
只有算术，枚举，指针，指向成员和允许类的类型参数。
Only arithmetic, enumeration, pointer, pointer to member, and class type arguments are allowed.
只有算术，枚举，指针，指向成员和允许类的类型参数。
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