复制构造函数和operator= 的区别

#include <iostream>
 
class Base
{
public:
    Base()
    {
        std::cout<<"constructor"<<std::endl;
    }
     
    Base(const Base &copy)
    {
        std::cout<<"copy constructor"<<std::endl;
    }
     
    const Base &operator=(Base &copy)
    {
        std::cout<<"operator="<<std::endl;
        return *this;
    }
    
};
 
int main(int argc, const char * argv[])
{
 
    Base a;     // 1
    Base b = a; // 2
    Base c(a);  // 3
     
    Base d;     // 4
    d = a;      // 5
     
    return 0;
}