有这样的代码：#include类 Base {public: Base(){ std::cout << \'构造函数 base\' << std::endl; } ~Base(){ std::cout << \'

有这样的代码：

#include <iostream>

class Base {
public:
    Base(){
        std::cout << "Constructor base" << std::endl;
    }
    ~Base(){
        std::cout << "Destructor base" << std::endl;
    }
    Base& operator=(const Base& a){
        std::cout << "Assignment base" << std::endl;
    }
};

class Derived : public Base{
public:

};

int main ( int argc, char **argv ) {
    Derived p;
    Derived p2;
    p2 = p;
    return 0;
}

经过g++4.6编译后的输出：

Constructor base
Constructor base
Assignment base
Destructor base
Destructor base

为什么虽然说赋值运算符不能被继承，但却调用基类的赋值运算符呢？