为什么指向成员函数的指针的模板参数推导失败？

Question

对于 g++ 5.4，这

struct B {
    void f() {}
}; 

struct D : public B {
    void g() {}
};

template <class T>
void foo(void (T::*)(), void (T::*)())
{}

int main()
{
    foo(&D::f, &D::g);
}

由于 "deduced conflicting types for parameter ‘T’ (‘B’ and ‘D’)" 而失败。为什么T不推导为D，完全匹配？

Answer 1

&D::f 的类型将是 void ( B::* )(void)

static_assert(::std::is_same<void ( B::* )(void), decltype(&D::f)>::value, "");
static_assert(::std::is_same<void ( D::* )(void), decltype(&D::f)>::value, ""); // error
static_assert(::std::is_same<void ( D::* )(void), decltype(&D::g)>::value, "");

这背后的基本原理是，否则即使 f 是B 的成员或比较 &D::f == &B::f.

作为解决方法，您可以执行 static_cast:

foo(static_cast<void (D::*)(void)>(&D::f), &D::g);

Answer 2

除了. The standard text in question, I believe, is at [expr.unary.op]/3，强调我的：

The result of the unary & operator is a pointer to its operand. The operand shall be an lvalue or a qualified-id. If the operand is a qualified-id naming a non-static or variant member m of some class C with type T, the result has type “pointer to member of class C of type T” and is a prvalue designating C::m.

你使用的限定id是D::f，但是它命名了B的一个成员函数（如果你需要我可以调出查找规则）。所以上一段中的class类型C，就是B。因此类型解析为 void ( B::* )(void).

为什么指向成员函数的指针的模板参数推导失败？

Why is template argument deduction failing for pointer-to-member-function?

c++

templates

template-argument-deduction