copy ctor 和 operator=

Question

#include <iostream>
#include <string>

class NoName { 
public: 
    NoName(): pstring(new std::string), i(0), d(0) {   
        std::cout << "default ctor is called" << std::endl; 
    }   

    NoName(const NoName& other) 
        : pstring(new std::string(*(other.pstring))), i(other.i), d(other.d) { 
            // C++ 默认的copy ctor是 
            // pstring(other.pstring)，这是浅拷贝，是错的 
            std::cout << "copy ctor is called" << std::endl; 
        }   

    NoName& operator=(const NoName rhs) { 
        std::cout << "operator= is called" << std::endl; 
        pstring = new std::string; 
        *pstring = *(rhs.pstring); 
        i = rhs.i; 
        d = rhs.d; 
        // C++ 默认的operator=是 
        // pstring = ths.pstring是浅拷贝，是错误的 
        return *this; 
    }   

private: 
    std::string *pstring; 
    int i; 
    double d; 
}; 

int main()
{
    std::cout << std::endl <<"------test------" << std::endl;
    std::cout << "NoName x, y:" << std::endl;
    NoName x, y;
    std::cout << std::endl;

    std::cout << "NoName z(x):" << std::endl;

    NoName z(x);
    std::cout << std::endl;

    std::cout << "x = y:" << std::endl;
    x = y;   
    std::cout << std::endl;

    return 0;
}

为什么x= y这一步即调用了copy ctor又调用了operator=

------test------
NoName x, y:
default ctor is called
default ctor is called

NoName z(x):
copy ctor is called

x = y:
copy ctor is called
operator= is called

Answer 1

赋值运算符的参数不是引用会导致调用拷贝构造函数，需要改成:

    NoName& operator=(const NoName& rhs) { 
        std::cout << "operator= is called" << std::endl; 
        pstring = new std::string; 
        *pstring = *(rhs.pstring); 
        i = rhs.i; 
        d = rhs.d; 
        // C++ 默认的operator=是 
        // pstring = ths.pstring是浅拷贝，是错误的 
        return *this; 
    }