开发者

Problem with cin.get() in C++?

I have the following code:

#include <conio.h>
using namespace std;
int _tmain(int argc, _TCHAR* argv[])
{
        int x = 0;
        cout << "Enter x: " ;
        cin >> x;
        if (cin.get() != '\n') // **line 1**
        {
            cin.ignore(1000,'\n');
            cout << "Enter number: ";
            cin >> x;
        }

        double y = 0;
        cout << "Enter y: ";
        cin >> y;       
        if (cin.get() != '\n'); // 开发者_Python百科**Line 2**
        {
            cin.ignore(1000,'\n');
            cout << "Enter y again: ";
            cin >> y;   
        }
        cout << x << ", " << y;

    _getch();

    return 0;
}

When executed, I can enter x value and it ignores Line 1 as I expected. However, when the program asks for y value, I inputed a value but the program did not ignore the while at Line 2? I don't understand, what is the difference between Line 1 and Line 2? And how can I make it work as expected?


if (cin.get() != '\n'); // **Line 2**
// you have sth here -^

Remove that semicolon. If it is there, the if statement basically does nothing.
Also, you're not testing wether the user really inputs a number... what if I input 'd' instead? :)

while(!(cin >> x)){
  // woops, something has gone wrong...
  // display a message to tell the user he made a mistake
  // and after that:
  cin.clear(); // clear all errors
  cin.ignore(1000,'\n'); // ignore until newline

  // and try again, while loop yay
}
// now we have correct input.
0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜