c++ - 如何使这个 operator= 重载工作？

Question

我正在尝试使operator=虚拟化，如下所示：

#include <iostream>
#include <string>

class Base {
public:
    std::string field1;
    Base(std::string str) : field1(str) {}

    virtual Base& operator=(const Base& other)
    {
        field1 = other.field1;
        return *this;
    }
};

class Derived : public Base {
public:
    std::string field2;
    Derived(std::string str1, std::string str2) : Base(str1), field2(str2) {}

    virtual Derived& operator=(const Derived& other) override
    {
        Base::operator=(other);
        field2 = other.field2;
        return *this;
    }
};

但是，这会产生编译器错误，因为该Derived函数实际上并没有重载任何东西，签名不同。

是否可以覆盖这样的operator=编写代码？

Base* ptr = new Derived("old 1", "old 2");
Derived derived("new 1", "new 2");

*ptr = derived; // <- use the derived class operator= to assign both field1 and field2

标签： c++overloadingvirtualoperator-keyword

Answer 1

该运算符

virtual Derived& operator=(const Derived& other);

不会覆盖基类中声明的复制赋值运算符。

你必须写

Derived& operator=(const Base& other) override;

即参数的类型应为const Base &。

这是一个演示程序。

#include <iostream>

struct A
{
    virtual A & operator =( const A & )
    {
        std::cout << "A::operator =\n";
        return *this;
    }
};

struct B : A
{
    virtual B & operator =( const A &a ) override
    {
        A::operator =( a );
        std::cout << "B::operator =\n";
        return *this;
    }
};


int main() 
{
    B b1;
    
    A &rb1 = b1;
    
    B b2;
    b2 = rb1;
    
    return 0;
}

它的输出是

A::operator =
B::operator =