第十九章 代码重用 5包含对系统的消耗

/*
//5 包含对系统的消耗
//由于book类的数据成员中包含量了3个String类的对像，因此在创建一个book类对像时也就不可避免的创建3个String类的对像，本节我们通过在String类和book类的构造和析构函数中添车输出语句演示这种消耗
#include "String.h"
class Book
{
public:
	Book();
	~Book(){ cout<<"Book类的析构函数执行...."<<endl;}
	Book(char*, char*, char*, float);
	//不能修改返回值，在函数内也不能修改，也不想调用复制构造函数，按地址传递
	const String& GetTitle()const{ return title; }
	const String& GetAuthor()const{ return author; }
	
	String& GetAuthor(){ return author; }

	const String& GetNumber()const{ return number; }
	float GetPrice()const{ return price; }

	void SetTitle(const String& s){ title = s; }
	void SetAuthor(const String& s){ author = s; }
	void SetNumber(const String& s){ number = s; }
	void SetPrice(float p){ price = p; }

	void SetTotal(const String&t, const String&a,const String&n, float p)
	{
	      title = t;
		  author = a;
		  number = n;
		  price = p;
	}
	
private:
    String title;  //书名
	String author; //作者
	String number; //编号
	float price;   //价格
};
//创建一本空图书
Book::Book():title(""),author(""),number(""),price(0){
	 cout<<"Book类的不带参数构造函数执行...."<<endl;
 
};
//创建一本有所有内容的图书
Book::Book(char* t, char* a, char* n, float p)
{
	cout<<"Book类的带参数构造函数执行...."<<endl;
	title = t; author=a; number = n; price=p;
}

int main()
{
	Book love("love","Jack","001",35.5);
	cout<<"书名:"<<love.GetTitle()<<endl;
	cout<<"作者:"<<love.GetAuthor()<<endl;
	cout<<"编号:"<<love.GetNumber()<<endl;
	cout<<"价格:"<<love.GetPrice()<<endl;
	love.SetPrice(55.5);
	cout<<"价格:"<<love.GetPrice()<<endl;

	love.SetTotal("hate","mak","002",39.9);
	cout<<"书名:"<<love.GetTitle()<<endl;
	cout<<"作者:"<<love.GetAuthor()<<endl;
	cout<<"编号:"<<love.GetNumber()<<endl;
	cout<<"价格:"<<love.GetPrice()<<endl;

	Book hoat;
	hoat = love;
	cout<<"书名:"<<hoat.GetTitle()<<endl;
	cout<<"作者:"<<hoat.GetAuthor()<<endl;
	cout<<"编号:"<<hoat.GetNumber()<<endl;
	cout<<"价格:"<<hoat.GetPrice()<<endl;
	if(love.GetNumber() == hoat.GetNumber()){
	   cout<<"两本书的编号相同"<<endl;
	}else{
	   cout<<"两本书的编号不相同"<<endl;
	}
	hoat.SetNumber("003");
	if(love.GetNumber() == hoat.GetNumber()){
	   cout<<"两本书的编号相同"<<endl;
	}else{
	   cout<<"两本书的编号不相同"<<endl;
	}

	//这里主要是
	//这是因为love.GetAuthor()和hoat.GetAuthor()函数返回的都是String类对像author，所以这会调用String类的operator+()函数来执行对两个对象的相加操作
	//由于GetAuthor()会返回一个const对像
	//const对像只能调用const成员函数
	//假如你不想修改String类的operator+()函数，那么还可以通过将book类的成员函数GetAuthor()重载为返回一个非常量对像的函数
	//第九行将GetAuthor()成员函数重载为一个返回非const对像的非const成员函数，这样，我们就可以执行String类非const对像的相加操作


	String loveAndHate = love.GetAuthor() + hoat.GetAuthor();
	Book LoveHoat;
	LoveHoat.SetAuthor(loveAndHate);
	LoveHoat.SetTitle("新书");
	cout<<"书名:"<<LoveHoat.GetTitle()<<endl;
	cout<<"作者名:"<<LoveHoat.GetAuthor()<<endl;
    return 0;
}
*/

　　

转载于:https://www.cnblogs.com/xiangxiaodong/archive/2012/09/24/2700893.html