函数重载对象层次结构

Question

在c ++中，如果你有以下

struct Base
{

};

struct Derived1 : Base
{

};

struct Derived2 : Derived1
{

};

struct Derived3 : Derived2
{
};

void f(const Derived2&)
{
    std::cout << "Derived2" << std::endl;
}

void f(const Derived1&)
{
    std::cout << "Derived1" << std::endl;
}

int main()
{
    f(Derived3());
    return 0;
}

f（Derived3（））调用“void f（const Derived2＆amp;）”是什么原因？是因为它在Derived3的对象层次结构中比Derived1更接近吗？

Answer 1

请参阅overload resolution rules了解c ++。

当尝试解析要调用的函数时，如果参数类型中没有直接匹配，则编译器必须对参数所需的转换进行排名，以便选择最佳候选者。

对于与继承层次结构中的对象相关的转换：

3) If Mid is derived (directly or indirectly) from Base, and Derived is derived (directly or indirectly) from Mid
a) Derived* to Mid* is better than Derived* to Base*
b) Derived to Mid& is better than Derived to Base&
c) Base::* to Mid::* is better than Base::* to Derived::*
d) Derived to Mid is better than Derived to Base
e) Mid* to Base* is better than Derived* to Base*
f) Mid to Base& is better than Derived to Base&
g) Mid::* to Derived::* is better than Base::* to Derived::*
h) Mid to Base is better than Derived to Base

案例b）适用于您的示例。