求简化~~~~

[西木凌萧]

// 程序内容 输入一数，判断是是否为数（包括各种类型的数）是 则输出，否则提示错误从新输入。最后提示是否据需输入Y则总新输入数 输入N结束 若输入其他字符 则提示错误从新输入。//

#include "stdafx.h"
#include <iostream>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
char f;
int n;


int i = 0;
while(true)
{
cout<<"please input n:"<<endl;
cin>>n;
if(cin.fail())
{
cout<<"data error!please input the integer:"<<endl;
cin.clear();
cin.sync();
}
else
{
cout<<n<<endl;
while(true)
{
cout<<"Whether to continue to run Y/N"<<endl;
cin>>f;

if(f!='Y'&&f!='N'&&f!='n'&&f!='y')
cout<<"data error!please input the integer:"<<endl;
else if(f == 'n' || f == 'N')
{
i = 1;
break;
}
else if(f == 'y' || f == 'Y')
break;
}

if(i == 1)
break;
}

}
cout<<"the system is over thank for using;"<<endl;
system( "Pause ");
return 0;
}

[测试NULL]

/* 帮你简化了一些 ~~ */
#include "stdafx.h"
#include <iostream>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
    char f;
    int n;
    
    do
    {
        cout<<"please input n:"<<endl;
        cin>>n;
        if(cin.fail())  
        {
            cout<<"data error!please input the integer:"<<endl;
            cin.clear();
            cin.sync();
            continue;
        }
        cout<<n<<endl;
        
        cout<<"Whether to continue to run Y/N"<<endl;
        cin>>f;
    }
    while(f == 'y' || f == 'Y');

    cout<<"the system is over thank for using;"<<endl;
    system( "Pause ");

    return 0;
}