关于重载的返回值问题

[wudichaoren2010]

class Object
{
private:
	int value;
public:
	Object(int i):value(i)
	{
	}

	const Object& operator=(const Object& obj)
	{
		value = obj.value;
		return *this;
	}

	const Object& operator=(const int i)
	{
		value = i;
		return *this;
	}

	int getValue(void) const
	{
		return value;
	}

	static void Test(void)
	{
		Object a(1),b(2),c(3);
		Widget x(1), y(2), z(3);

		a  = b = c = 4;
		//((a=b)=c)=4;		// Error	2	error C2678: binary '=' :/
							no operator found which takes a left-hand operand of type 'const Object' /
							(or there is no acceptable conversion)	
		
		
	}

const Object& operator=(const int i)
	{
		value = i;
		return *this;
	}

返回值是是什么？
是引用还是地址或者原对象？

[encoderlee]

返回值是Object的引用 比如 Object ob(10); ob = 5; //该表达式返回的是ob的const引用

[wudichaoren2010]

class Base
{private:
int Y;
public:
Base(int y=0) {Y=y;cout<<"Base("<<y<<")\n";}
~Base() {cout<<"~Base()\n";}
void print() {cout <<Y<< "";}
};
class Derived:public Base
{
private:
	int Z;
public:
	Derived(int y, int z):Base(y)
	{
		Z=z;
		cout<<"Derived("<<y<<","<<z<<")\n";
	}
	~Derived() {cout<<"～Derived()\n";}
void print()
{
	Base::print();
	cout<<Z<<endl;
}
};
void main()
{Derived d(10,20);
d.print();
}

为什么不输出Base(0);

[encoderlee]

Derived d(10,20); 你构造Derived对象的时候传给y的参数是10, Derived(int y, int z):Base(y) Base的构造函数接收到的参数也是10 当你传入一个参数10给它后，Base(int y=0)该默认参数0就不起作用了

[wudichaoren2010]

谢谢