c++ - 如何使operator=像operator+一样接受参数的导数？

Question

我不明白为什么a = b不像operator+那样打印出来自operator= 5的值(这似乎允许导数)。为什么它这样做而不是允许衍生品，我怎样才能让它也接受衍生品？

#include <iostream>

class A {
public:
    int value = 5;
};

class B : public A {
public:
    void operator=(const A& rhs) {
        std::cout << "value from operator= " << rhs.value << std::endl;
    }
    
    void operator+(const A& rhs) {
        std::cout << "value from operator+ " << rhs.value << std::endl;
    }
};

int main() {
    
    B a;
    B b;
    A c;
    
    std::cout << "testing operator=" << std::endl;
    std::cout << "testing with B:" << std::endl;
    a = b; // doesn't print anything
    std::cout << "testing with A:" << std::endl;
    a = c; // outputs: value from operator= 5
    
    std::cout << std::endl;
    
    std::cout << "testing operator+" << std::endl;
    std::cout << "testing with B:" << std::endl;
    a + b; // outputs: value from operator+ 5
    std::cout << "testing with A:" << std::endl;
    a + c; // outputs: value from operator+ 5
    
}

Answer 1

B::operator=(const A& rhs) 不是复制赋值或移动赋值运算符，因此 implicitly-defined copy assignment operator仍然存在，并且有签名

B& operator=(const B&);

当执行a = b时，这个运算符在重载解析中胜过你的，因为调用它时，不需要派生到基数的转换B -> A，与赋值运算符不同。

让 B 和 A 使用此运算符的一种方法是:

B& operator=(const A& rhs) {
    std::cout << "value from operator= " << rhs.value << std::endl;
    return *this;
}

B& operator=(const B& rhs) {
    return static_cast<A&>(*this) = rhs;
}

更传统的是，A 应该有自己的用户定义的复制赋值运算符，而 B 将通过 return A::operator=(rhs) 转发给它;