为什么此typedef允许我在此模板中使用基类指向成员函数的指针？

Question

If I attempt to compile the following code in MSVC:

template <typename DELEGATE>
void newButton(DELEGATE *obj, int (DELEGATE::*method)(int))
{
    std::function<int(int)> callback = std::bind(
        method, obj, std::placeholders::_1);
    // ...
}

class Base
{
public:
    virtual int test(int f) { return f * f; }
};

class Derived : public Base
{
};

int main()
{
    Derived d;
    newButton(&d, &Base::test);
}

I get a compiler error:

'void newButton(DELEGATE *,int (__thiscall DELEGATE::* )(int))' : 
 template parameter 'DELEGATE' is ambiguous
          could be 'Base'
          or       'Derived'

This is reasonable. The template expects an identical type for obj and method , but they are not exactly the same.

BUT, if I replace the pointer-to-member-function declaration with this template struct typedef, it compiles!

template <typename DELEGATE>
struct ButtonAction
{
    typedef int (DELEGATE::*Type)(int);
};

template <typename DELEGATE>
void newButton(DELEGATE *obj, typename ButtonAction<DELEGATE>::Type method)
{
    std::function<int(int)> callback = std::bind(
        method, obj, std::placeholders::_1);
    // ...
}

// rest same as before

Why? I would have expected the typedef to get resolved down to the exact same pointer-to-member-function type, and cause the same template error.

Answer 1

The reason is that in ButtonAction<DELEGATE>::Type , DELEGATE appears in a non-deduced context - the compiler cannot deduce DELEGATE from this, so it does not try. Therefore, the deduction is performed from the first argument only, and so it's unambiguous.

As to why DELEGATE cannot be deduced in this context - try to imagine what the process would need to be: inspect ButtonAction<T> for every possible type T and compare its nested typedef Type against the argument type. And note that there are infinitely many possible types.

A rule of thumb is: everything to the left of :: is a non-deduced context.

Answer 2

在第二个示例中， method参数不参与类型推导，因此将DELEGATE推导为Derived ，并将&Base::test隐式转换为(Derived::*)(int) 。