c++ - 从指向基类成员的指针进行模板推导

Question

我不知道我做错了什么。从指向基类成员的指针进行模板推导时出现问题 - &DerivedClass::BaseClassMemeber。

完整示例:

#include <vcl.h>
#include <tchar.h>

struct Base
{
   int BaseClassMember;
};

struct Derived : public Base
{
};

template<class T1, class T2>
void Test(T1& Param1, T2 T1::* Param2)
{
}

int _tmain()
{
   Derived inst;
   // Compile error E2285 Could not find a match for 'Test<T1,T2>(B,int A::*) - BCC32
   // Error 1 error C2782: 'void Test(T1 &,T2 T1::* )' : template parameter 'T1' is   ambiguous - MS VS8
   Test(inst, &Derived::BaseClassMember);

   // Works great
   Test<Derived>(inst, &Derived::BaseClassMember);
   return 0;
 }

我可以找到几种解决方法，例如额外的测试函数重载带有一个模板参数、static_cast、隐式部分特化(测试)。

但我对为什么编译器不能使用 &DerivedClass::BaseClassMemeber 中明确指定的类的原因感兴趣。就是那个问题。如果您有更优雅的问题解决方案，欢迎。

Answer 1

&Derived::BaseClassMember 的类型为 int Base::*，而不是 int Derived::*。引用标准:

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 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. [... skip ...] [Example:

struct A { int i; };
struct B : A { };
... &B::i ...             // has type int A::*i

— end example]

如果您需要此类型，则必须将值转换为 int Derived::*。