c++ - 为什么具有相同名称但不同签名的多个继承函数不会被视为重载函数？

Question

以下代码段在编译期间会产生“对 foo 的模糊调用”错误，我想知道是否有任何方法可以在不完全限定对 foo 的调用的情况下解决此问题:

#include <iostream>

struct Base1{
    void foo(int){
    }
};

struct Base2{
    void foo(float){
    }
};

struct Derived : public Base1, public Base2{
};

int main(){
    Derived d;
    d.foo(5);

    std::cin.get();
    return 0;
}

所以，问题正如标题所说。想法？我的意思是，以下工作完美无缺:

#include <iostream>

struct Base{
    void foo(int){
    }
};

struct Derived : public Base{
    void foo(float){
    }
};

int main(){
    Derived d;
    d.foo(5);

    std::cin.get();
    return 0;
}

Answer 1

成员查找规则在第 10.2/2 节中定义

The following steps define the result of name lookup in a class scope, C. First, every declaration for the name in the class and in each of its base class sub-objects is considered. A member name f in one sub-object B hides a member name f in a sub-object A if A is a base class sub-object of B. Any declarations that are so hidden are eliminated from consideration. Each of these declarations that was introduced by a using-declaration is considered to be from each sub-object of C that is of the type containing the declara-tion designated by the using-declaration. If the resulting set of declarations are not all from sub-objects of the same type, or the set has a nonstatic member and includes members from distinct sub-objects, there is an ambiguity and the program is ill-formed. Otherwise that set is the result of the lookup.

class A {
public:
  int f(int);

};
class B {
public:
   int f();

};
class C : public A, public B {};
int main()
{
     C c;
     c.f(); // ambiguous
}

所以你可以使用 using 声明 A::f 和 B::f 来解决这种歧义

class C : public A, public B {
     using A::f;
     using B::f;

};

int main()
{
     C c;
     c.f(); // fine
}

第二个代码完美无缺，因为 void foo(float) 在 C 的范围内。实际上 d.foo(5); 调用 void foo(float) 而不是 int 版本。