滴水逆向_引用_友元函数_运算符重载

作业：

运算符号重载实现。

struct Person
{
public:
	int x;
	int y;
public:
	Person()
	{
		this->x = 10;
		this->y = 20;
	}
	 Person(int x, int y)
	{
		this->x = x;
		this->y = y;
	}

	//申明友元函数
	 void  Printf(const Person& p)
	 {
		 printf("%d  %d",p.x,p.y);
	}
	//友元函数重载

    Person operator + (const Person& p);
	Person operator - (const Person& p);
	Person operator * (const Person& p);
	Person operator / (const Person& p);
	bool operator >(const Person& p);
	bool operator <(const Person& p);
	bool operator >=(const Person& p);
	bool operator <=(const Person& p);
	bool operator ==(const Person& p);

};


Person Person:: operator + (const Person & p )
{
	this->x =this->x +  p.x;
	this->y = this->y + p.y;

	return *this;
}
Person Person:: operator - (const Person& p)
{
	this->x = this->x - p.x;
	this->y = this->y - p.y;

	return *this;
}
Person Person:: operator * (const Person& p)
{
	this->x = this->x * p.x;
	this->y = this->y * p.y;

	return *this;
}
Person Person:: operator / (const Person& p)
{
	this->x = this->x / p.x;
	this->y = this->y / p.y;
	return *this;
}

bool Person ::operator >(const Person& p)
{
	if (this->x > p.x && this->y > p.y)
	{
		return true;
	}
	return false;
}
bool Person ::operator <(const Person& p)
{
	if (this->x < p.x && this->y< p.y)
	{
		return true;
	}
	return false;
}
bool Person ::operator >=(const Person& p)
{
	if (this->x >= p.x && this->y >= p.y)
	{
		return true;
	}
	return false;
}
bool Person ::operator <=(const Person& p)
{
	if (this->x <= p.x && this->y <= p.y)
	{
		return true;
	}
	return false;
}
bool Person ::operator ==(const Person& p)
{
	if (this->x == p.x && this->y == p.y)
	{
		return true;
	}
	return false;
}

2 引用和指针的区别

	x = (int*)10;
00592EB1  mov         dword ptr [x],0Ah

指针修改指向生成的反汇编代码。

引用

	x = 10;
00591A01  mov         eax,dword ptr [x]  
00591A04  mov         dword ptr [eax],0Ah  

引用是不可能出现修改指向的反汇编代码的。

这也就是反汇编中唯一能看出来的瑕疵。