一道关于 swith（） 的问题 谢谢~！

[ufdxkm]

#include <iostream>
const int charsize=20;
const int menber=3;

struct bop{
	char fullname[charsize];
	char title[charsize];
	char bopname[charsize];
	int preference;
};
int main()
{
	using namespace std;
	bop person[menber]=
	{
	
		{
			"pengchang",
				"go away",
				"Crius.Zc",
				3,
		},{
		
			"pengruze",
			"come and lat's go",
			"Rz",
			2,
		},{
			"unkown",
				"i don't know",
				"zz",
				1,
			}
	};

    cout << "Benevolent Order of Programmers Report\n";
    cout << "a. display by name\t    b. dispaly by title\n"
            "c. display by bopname\t   d. display by preference\n"
            "q. quit\n";
	
	char ch;
	cout<<"Enter you choice :";
	while(cin>>ch &&ch !='q' &&ch !='Q')
	{
		while(!(ch>='a' && ch <='d' || ch>='A' && ch<='D'))
		{
			while(cin.get() !='\n')
				continue;
			cout<<"That is not a choice ,Choice again:";
			cin>>ch;
			
		}
		switch(ch)
		{
		case'a':
		case'A':for(int i=0;i<menber;i++)
					cout<<person[i].fullname<<endl;break;
		case'b':
		case'B':for(int i=0;i<menber;i++)
					cout<<person[i].title<<endl;break;
		case'c':
		case'C':for(int i=0;i<menber;i++)
					cout<<person[i].bopname<<endl;break;
		case'd':
		case'D':for(int i=0;i<menber;i++)
				{
					switch(person[i].preference)
					{
					case 0:cout<<person[i].fullname<<endl;break;
					case 1:cout<<person[i].title<<endl;break;
					case 2:cout<<person[i].bopname<<endl;break;
					}
				}break;

		}
		while(cin.get()!='\n')
			continue;
		cout<<"next choice: ";
		cin>>ch;
	}
	cout<<"bye~!"<<endl;
		
	
	return 0;
}

就是这个 连接 http://topic.csdn.net/u/20070923/18/77cd4bf8-4a81-403a-ad60-b7ab42d869b5.html 里的例子
如果那个switch
case 'q':
case 'Q': cout << "Bye!\n"; exit(EXIT_FAILURE);
不用exit(EXIT_FAILURE); 还可以用什么办法 输入q 就可以退出整个程序
我上边的那个例子如果第一次输入（假如一个）x,会显示
That is not a choice ,Choice again:
然后再输入q，还是输出这句话
因为执行这个循环
while(!(ch>='a' && ch <='d' || ch>='A' && ch<='D'))
{
while(cin.get() !='\n')
continue;
cout<<"That is not a choice ,Choice again:";
cin>>ch;

}

所以 有什么办法可以在输入x后输入q也可以退出？


谢谢大家！

[sagegz]

现在看来也只能用goto了,不过我只会汇编里的JMP
等待牛人了.

[9527-]

#include <iostream>
using namespace std;

const int charsize=20;
const int menber=3;

struct bop{
char fullname[charsize];
char title[charsize];
char bopname[charsize];
int preference;
};
int main(){
bop person[menber]={{"pengchang","go away","Crius.Zc",3},{"pengruze","come and lat's go","Rz",2},{"unkown","i don't know","zz",1}};
cout << "Benevolent Order of Programmers Report\n";
cout << "a. display by name\t b. dispaly by title\n"
"c. display by bopname\t d. display by preference\n"
"q. quit\n";
char ch;
cout<<"Enter you choice :";
cin>>ch;
while(ch !='q' && ch !='Q'){
while(!(ch>='a' && ch <='d' || ch>='A' && ch<='D')){
while(cin.get() !='\n')
continue;
cout<<"That is not a choice ,Choice again:";
cin>>ch;
if(ch=='q' || ch== 'Q') //这里加2行
return( 0 ); //exit(-1);//这样做是你要的效果吗 }
switch(ch){
case'a':
case'A':for(int i=0;i<menber;i++)
cout<<person[i].fullname<<endl;break;
case'b':
case'B':for(int i=0;i<menber;i++)
cout<<person[i].title<<endl;break;
case'c':
case'C':for(int i=0;i<menber;i++)
cout<<person[i].bopname<<endl;break;
case'd':
case'D':for(int i=0;i<menber;i++)
{
switch(person[i].preference)
{
case 0:cout<<person[i].fullname<<endl;break;
case 1:cout<<person[i].title<<endl;break;
case 2:cout<<person[i].bopname<<endl;break;
}
}break;
}
while(cin.get()!='\n')
continue;
cout<<"next choice: ";
cin>>ch;
}
cout<<"bye~!"<<endl;
return 0;
}

[xuruichen]

[Quote=引用 9 楼 waizqfor 的回复:]
C/C++ code
while(!(ch>='a' && ch <='d' || ch>='A' && ch <='D'))
{
while(cin.get() !='\n')
continue;
cout < <"That is not a choice ,Choice again:";
cin>>ch;

if(ch=='q'||ch=='Q')//这里加个判断 是不是q然后推出程序
exit(1);

}
[/Quote]
correct

[tankinc]

if(ch=='q'||ch=='Q')//这里加个判断 是不是q然后推出程序
exit(1);
这是最省力的方法。

[barbara2008]

直接强制return

[barbara2008]

while(!(ch>='a' && ch <='d' || ch>='A' && ch<='D'))
{
while(cin.get() !='\n')
continue;
cout<<"That is not a choice ,Choice again:";
cin>>ch;
//这里加个判断，跳出或者饭返回

}

[ufdxkm]

强制return
最后那句 bye~! 不能输出了阿

[barbara2008]

强制return和你开玩笑的，呵呵

[ufdxkm]

加break 调处？ 还是跳出一层循环阿。
还是执行那个循环。。

[sagegz]

帮你+了2行代码,现在可以解决输入X后再输入Q不能结束程序的问题了.

#include <iostream>
using namespace std;

const int charsize=20;
const int menber=3;

struct bop{
    char fullname[charsize];
    char title[charsize];
    char bopname[charsize];
    int preference;
};
int main(){    
    bop person[menber]={{"pengchang","go away","Crius.Zc",3},{"pengruze","come and lat's go","Rz",2},{"unkown","i don't know","zz",1}};
    cout << "Benevolent Order of Programmers Report\n";
    cout << "a. display by name\t    b. dispaly by title\n"
            "c. display by bopname\t   d. display by preference\n"
            "q. quit\n";    
    char ch;
    cout<<"Enter you choice :";
	cin>>ch;
    while(ch !='q' && ch !='Q'){
        while(!(ch>='a' && ch <='d' || ch>='A' && ch<='D')){
            while(cin.get() !='\n')
                continue;
            cout<<"That is not a choice ,Choice again:";
            cin>>ch;
			if(ch=='q' || ch== 'Q')	//这里加2行
				exit(-1);
        }
		switch(ch){
		case'a':
		case'A':for(int i=0;i<menber;i++)
                    cout<<person[i].fullname<<endl;break;
        case'b':
        case'B':for(int i=0;i<menber;i++)
                    cout<<person[i].title<<endl;break;
        case'c':
        case'C':for(int i=0;i<menber;i++)
                    cout<<person[i].bopname<<endl;break;
        case'd':
        case'D':for(int i=0;i<menber;i++)
                {
                    switch(person[i].preference)
                    {
                    case 0:cout<<person[i].fullname<<endl;break;
                    case 1:cout<<person[i].title<<endl;break;
                    case 2:cout<<person[i].bopname<<endl;break;
                    }
                }break;
        }
        while(cin.get()!='\n')
            continue;
        cout<<"next choice: ";
        cin>>ch;
    }
    cout<<"bye~!"<<endl;    
    return 0;
}

[向良玉]

正确的[Quote=引用 6 楼 sagegz 的回复:]
帮你+了2行代码,现在可以解决输入X后再输入Q不能结束程序的问题了.

C/C++ code
#include <iostream>
using namespace std;

const int charsize=20;
const int menber=3;

struct bop{
char fullname[charsize];
char title[charsize];
char bopname[charsize];
int preference;
};
int main(){
bop person[menber]={{"pengchang","go away","Crius.Zc",3},{"pengruze","come and lat…
[/Quote]

[向良玉]

正确的[Quote=引用 6 楼 sagegz 的回复:]
帮你+了2行代码,现在可以解决输入X后再输入Q不能结束程序的问题了.

C/C++ code
#include <iostream>
using namespace std;

const int charsize=20;
const int menber=3;

struct bop{
char fullname[charsize];
char title[charsize];
char bopname[charsize];
int preference;
};
int main(){
bop person[menber]={{"pengchang","go away","Crius.Zc",3},{"pengruze","come and lat…
[/Quote]

[waizqfor]

while(!(ch>='a' && ch <='d' || ch>='A' && ch <='D')) 
{ 
while(cin.get() !='\n') 
continue; 
cout < <"That is not a choice ,Choice again:"; 
cin>>ch; 

if(ch=='q'||ch=='Q')//这里加个判断 是不是q然后推出程序
exit(1);

} 

[xiaoyisnail]

[Quote=引用 6 楼 sagegz 的回复:]
帮你+了2行代码,现在可以解决输入X后再输入Q不能结束程序的问题了.
C/C++ code#include<iostream>usingnamespacestd;constintcharsize=20;constintmenber=3;structbop{charfullname[charsize];chartitle[charsize];charbopname[charsize];intpreference;
};intmain(){
bop person[menber]={{"pengchang","go away","Crius.Zc",3},{"pengruze","come and lat's go","Rz",2},{"unkown","i don't know","zz",1}};
cout…
[/Quote]

up

[ufdxkm]

谢谢各位拉
帮顶的就不加分拉 就20分 不好意思拉~。。。
我看的C++ Primer..PLus只是还没有见过exit()函数，而且我觉得这道题应该不是考exit这个函数的。。
所以希望下面的朋友是否可以不用exit() 作出来。。。 嘻嘻。。
其实我是想学点东西，不是只为作这道题的哈~

[ufdxkm]

大家讨论一下哈

[waizqfor]

[Quote=引用 11 楼 eidolondedidi 的回复:]
谢谢各位拉
帮顶的就不加分拉 就20分 不好意思拉~。。。
我看的C++ Primer..PLus只是还没有见过exit()函数，而且我觉得这道题应该不是考exit这个函数的。。
所以希望下面的朋友是否可以不用exit() 作出来。。。 嘻嘻。。
其实我是想学点东西，不是只为作这道题的哈~
[/Quote]
那就用 GO TO 呵呵