为什么重载赋值运算符不会被继承？

Question

为什么这段代码:

class X {
public:
    X& operator=(int p) {
        return *this;
    }
    X& operator+(int p) {
        return *this;
    }
};

class Y : public X { };

int main() {
    X x;
    Y y;
    x + 2;
    y + 3;
    x = 2;
    y = 3;
}

给出错误:

prog.cpp: In function ‘int main()’:
prog.cpp:14:9: error: no match for ‘operator=’ in ‘y = 3’
prog.cpp:14:9: note: candidates are:
prog.cpp:8:7: note: Y& Y::operator=(const Y&)
prog.cpp:8:7: note:   no known conversion for argument 1 from ‘int’ to ‘const Y&’
prog.cpp:8:7: note: Y& Y::operator=(Y&&)
prog.cpp:8:7: note:   no known conversion for argument 1 from ‘int’ to ‘Y&&’

为什么+运营商继承了,但=运营商没有？

Answer 1

类Y包含隐式声明的赋值运算符,它隐藏在基类中声明的运算符.通常,在派生类中声明函数会隐藏在基类中声明的具有相同名称的任何函数.

如果您想同时使用Y,请使用using声明:

class Y : public X {
public:
    using X::operator=;
};