令operator=返回一个reference to *this

int main()
{int a, b, c = 5;a = b = c;cout << a;
}

这个代码，很明显输出的是5。所以我们在写这种连续赋值的时候，其对应的赋值运算符应当返回一个*this ：

class A
{
public:A(string ss, int x) :s(ss), a(x) {};A& operator=(const A& aa){this->a = aa.a;this->s = aa.s;return *this;}friend ostream& operator<<(ostream& out,A& aa);private:string s;int a;
};
ostream& operator<<(ostream& out,A& aa)
{out << aa.a<<" "<<aa.s;return out;
}
int main()
{A a("a", 1);A b("b", 2);A c("c", 3);a = b = c;cout << a << endl;cout << b << endl;cout << c << endl;
}

结果符合预期！