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！