将一个指向成员的指针函数转换为同一个类的另一个指针

Question

Is it legal to cast a pointer-to-member-function to another pointer-to-member-function of the same class using reinterpret_cast ? The following example works. But is it legal?

#include<iostream>
#include<vector>
#include<string>

class A
{
    public:
    void func_int(int x) { std::cout << x << std::endl; }
    void func_string(std::string const& x) { std::cout << x << std::endl; }
};

int main()
{
    std::vector<void(A::*)()> v;
    v.push_back(reinterpret_cast<void(A::*)()>(&A::func_int));
    v.push_back(reinterpret_cast<void(A::*)()>(&A::func_string));
    A a;
    (a.*reinterpret_cast<void(A::*)(int)>(v[0]))(5);
    (a.*reinterpret_cast<void(A::*)(std::string const&)>(v[1]))(std::string{"Test"});

    return 0;
}

Answer 1

At [expr.reinterpret.cast] the C++ draft states:

A prvalue of type “pointer to member of X of type T1 ” can be explicitly converted to a prvalue of a different type “pointer to member of Y of type T2 ” if T1 and T2 are both function types or both object types. ⁷² The null member pointer value ( [conv.mem] ) is converted to the null member pointer value of the destination type. The result of this conversion is unspecified, except in the following cases:

• converting a prvalue of type “pointer to member function” to a different pointer to member function type and back to its original type yields the original pointer to member value.

• converting a prvalue of type “pointer to data member of X of type T1 ” to the type “pointer to data member of Y of type T2 ” (where the alignment requirements of T2 are no stricter than those of T1 ) and back to its original type yields the original pointer to member value.

72) T1 and T2 may have different cv -qualifiers, subject to the overall restriction that a reinterpret_cast cannot cast away constness.

Since you are converting your "pointer to member function" to a different "pointer to member function" type and back, it yields the original value. This is both legal and well-defined behaviour. So your code should work properly.