c++改错题

[guocai_yao]

//要求：不改变程序结构，在出错行的行首做一标记*，并将该多最少的正确语句下载所在行下放或者出错部位的右侧
#include<iostream>
using namespace std;

class A
{
	int i;
public:
	A(int x = 0):i(x){}
	virtual int get(){return i;}
	//
	//
};

class B:public A
{
public:
	B(int x = 1):A(x){}
	void operator!(){cout<<"i = "<<get()<<endl;}
	//
        //
};
class C:public B
{
	int i;
public:
	C(int x = 2, int y):B(x),i(y){}
	int operator+=(A &x){return i+=x.get();}
	int get(){return i;}
};

int main()
{
	A &p1 = *(A *)new B();
	A &p2 = *(A *)new C(10);
	!*p1;
	cout<<"p2{i}="<<(p2+=p1)<<endl;
	cout<<"p2{i}="<p2.get()<<endl;
	return 0;
}

//新手，谢谢

[xiaoyisnail]

看不懂你的要求，也看不懂你的代码想做什么

[guocai_yao]

就是这段程序有错误;//地方缺少语句，改完后程序能运行就行

//要求：不改变程序结构，在出错行的行首做一标记*，并将该多最少的正确语句下载所在行下放或者出错部位的右侧
#include<iostream>
using namespace std;

class A
{
    int i;
public:
    A(int x = 0):i(x){}
    virtual int get(){return i;}
    //
    //
};

class B:public A
{
public:
    B(int x = 1):A(x){}
    void operator!(){cout<<"i = "<<get()<<endl;}
    //
        //
};
class C:public B
{
    int i;
public:
    C(int x = 2, int y):B(x),i(y){}
    int operator+=(A &x){return i+=x.get();}
    int get(){return i;}
};

int main()
{
    A &p1 = *(A *)new B();
    A &p2 = *(A *)new C(10);
    !*p1;//要求显示的结果为i = 1;
    cout<<"p2{i}="<<(p2+=p1)<<endl;//要求显示结果为p2{i} = 11;
    cout<<"p2{i}="<p2.get()<<endl;//要求显示结果为p2{i} = 11;
    return 0;
}

[elegant87]

int main()
{
    A &p1 = *(A *)new B();
    C &p2 = *(C *)new C();
    !&p1;
    cout<<"p2{i}="<<(p2+=p1)<<endl;
    cout<<"p2{i}="<<p2.get()<<endl;
    system("pause");
    return 0;
}

[帅得不敢出门]

#include<iostream>
using namespace std;

class A
{
    int i;
public:
    A(int x = 0):i(x){}
    virtual int get(){return i;}
    //
    //
};

class B:public A
{
public:
    B(int x = 1):A(x){}
    void operator!(){cout<<"i = "<<get()<<endl;}
    //
        //
};
class C:public B
{
    int i;
public:
    C(int x = 2):B(x),i(x){}
    int operator+=(A &x){return i+=x.get();}
    int get(){return i;}
};

int main()
{
    B *p1 = new B();
    C *p2 = new C(10);
    !*p1;
	*p2+=*p1;
    cout<<"p2{i}="<<p2->get() << endl;
    return 0;
}

[guocai_yao]

楼上的改后还是报错

[帅得不敢出门]

错误比较多，
不改结构是不行的。
使用上也有问题。

[guocai_yao]

    cout<<"p2{i}="<<(p2+=p1)<<endl;//要求显示结果为p2{i} = 11;
    cout<<"p2{i}="<p2.get()<<endl;//要求显示结果为p2{i} = 11;

楼上的程序结构已经发生了改变

    *p2+=*p1;

[ltc_mouse]

不知道算不算改动最少,修改放在了屏蔽的代码下方

class C:public B
{
    int i;
public:
    ///C(int x = 2, int y):B(x),i(y){}
    C(int y, int x = 2):B(x),i(y){}
    int operator+=(A &x){return i+=x.get();}
    int get(){return i;}
};

int main()
{
    A &p1 = *(A *)new B();
    A &p2 = *(A *)new C(10);
    ///!*p1;
    !( dynamic_cast<B&>(p1) );

    ///cout<<"p2{i}="<<(p2+=p1)<<endl;
    cout<<"p2{i}="<<(dynamic_cast<C&>(p2)+=p1)<<endl;
    cout<<"p2{i}="<p2.get()<<endl;
    return 0;
}

[guocai_yao]

#include<iostream>
using namespace std;

class A
{
	int i;
public:
	A(int x = 0):i(x){}
	virtual int get(){return i;}
	virtual void operator!() = 0;
	virtual int operator+=() = 0;
};

class B:public A
{
public:
	B(int x = 1):A(x){}
	void operator!(){cout<<"i = "<<get()<<endl;}
	int operator+=(A &x){return i+=x.get();}
};
class C:public B
{
	int i;
public:
	C(int y, int x = 2):B(x),i(y){}
	int operator+=(A &x){return i+=x.get();}
	int get(){return i;}
};

int main()
{
	A &p1 = *(A *)new B();
	A &p2 = *(A *)new C(10);
	!(B&)p1;
	cout<<"p2{i}="<<(((C&)p2)+=p1)<<endl;
	cout<<"p2{i}="<<(C&)p2.get()<<endl;
	return 0;
}

有同学这样改的，也是不对，大概这个意思

[guocai_yao]

#include<iostream>
using namespace std;

class A
{
    int i;
public:
    A(int x = 0):i(x){}
    virtual int get(){return i;}
    //还要加代码
    //还要加代码
};

class B:public A
{
public:
    B(int x = 1):A(x){}
    void operator!(){cout<<"i = "<<get()<<endl;}
    //还要加代码
    //还要加代码
};
class C:public B
{
    int i;
public:
    C(int x = 2, int y):B(x),i(y){}
    int operator+=(A &x){return i+=x.get();}
    int get(){return i;}
};

int main()
{
    A &p1 = *(A *)new B();
    A &p2 = *(A *)new C(10);
    !*p1;//要求显示的结果为i = 1;
    cout<<"p2{i}="<<(p2+=p1)<<endl;//要求显示结果为p2{i} = 11;
    cout<<"p2{i}="<p2.get()<<endl;//要求显示结果为p2{i} = 11;
    return 0;
}

[xiaoyisnail]

#include<iostream> 
using namespace std; 
class A 
{ 
	int i; 
public: 
	A(int x = 0):i(x){} 
	virtual int get(){return i;} 
	virtual void operator!(){}
	virtual int operator+=(A &x){return 0;}
}; 
class B:public A 
{ 
public: 
	B(int x = 1):A(x){} 
	void operator!(){cout<<"i = "<<get()<<endl;} 
	//  
	// 
}; 
class C:public B 
{ 
	int i; 
public: 
	C(int x = 2, int y=0):B(x),i(y){} 

	int operator+=(A &x)
	{
		return i+=x.get();
	} 
	int get()
	{
		return i;
	} 
}; 
		
		
int main() 
{ 
	A &p1 = *(A *)new B(); 
	A &p2 = *(A *)new C(10, 10); 
	!p1;//要求显示的结果为i = 1;  
	cout<<"p2{i}="<<(p2+=p1)<<endl;//要求显示结果为p2{i} = 11;  
	cout<<"p2{i}="<<p2.get()<<endl;//要求显示结果为p2{i} = 11;  
	return 0; 
}

[xiaoyisnail]

#include<iostream> 
using namespace std; 
class A 
{ 
	int i; 
public: 
	A(int x = 0):i(x){} 
	virtual int get(){return i;} 
	virtual A& operator*(){return *this;}
	virtual void operator!(){}
	virtual int operator+=(A &x){return i+=x.get();}
}; 
class B:public A 
{ 
public: 
	B(int x = 1):A(x){} 
	void operator!(){cout<<"i = "<<get()<<endl;} 
	B& operator*(){return *this;}  
}; 
class C:public B 
{ 
	int i; 
public: 
	C(int x = 2, int y):B(x),i(y){} 

	int operator+=(A &x)
	{
		return i+=x.get();
	} 
	int get()
	{
		return i;
	} 
}; 
		
		
int main() 
{ 
	A &p1 = *(A *)new B(); 
	A &p2 = *(A *)new C(10, 10); 
	!*p1;//要求显示的结果为i = 1;  
	cout<<"p2{i}="<<(p2+=p1)<<endl;//要求显示结果为p2{i} = 11;  
	cout<<"p2{i}="<<p2.get()<<endl;//要求显示结果为p2{i} = 11;  
	return 0; 
}

[bfhtian]

mark

[lnever]

呵呵，楼主的意思很多人没有看清。
要声明借口或者重载运算符。
照楼主最后的代码，
应该是A声明＋重载运算符，和get函数，
B重载＋运算符和get函数。

[guocai_yao]

12楼正解

[guocai_yao]

对不起大家了，等我分挣够了再来结贴哈囧

[speeder53]

mk