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我创建了一个程序，使用 Scanner 从用户那里获取一个数字，并在它是 1 到 100 的整数时将其保存到“a”。请参阅下面的 Java 文件：

public class Parity_Check {
  private static Scanner sc;

  public static void main(String[] args) throws InterruptedException {

    sc = new Scanner(System.in);
    int a, b;
    System.out.print("Enter a number     between 1 and 100: ");

    while(true) {
      b = 0;
      if(!sc.hasNextInt()) {
        System.out.print("That isn't an integer! Try again: "); 
        sc.next();
      }
      else{
        b = sc.nextInt();
        if(b < 1 || b > 100) {
          System.out.print("That integer isn't between 1 and 100! Try again: "); 
          sc.next();
        }
        else{
          a = b; 
          break;
        }
      }
    }
    System.out.print("The number is: "+a+".");
  }
}

我遇到的问题如下：程序返回“那个整数不在 1 到 100 之间！再试一次：，“它等待来自用户的两个输入（而不是它应该输入的那个）——第一个被完全忽略！这是我为说明问题而运行的控制台会话：

"Enter a number between 1 and 100: 2.5
That isn't an integer! Try again: 101
That integer isn't between 1 and 100! Try again: Apple.
42
The number is: 42.”

正如你所看到的，它甚至没有注意到输入"Apple". 我完全迷失了为什么这不能正常工作，就像这样：

"Enter a number between 1 and 100: 2.5
That isn't an integer! Try again: 101
That integer isn't between 1 and 100! Try again: Apple.
That isn't an integer! Try again: 42
The number is: 42.”

我对Java很陌生，所以一个很好解释的答案将是天赐之物；我对它为什么不起作用比如何修复它更感兴趣，因为希望我能够学习。

顺便说一句，我正在使用最新版本的 Eclipse 64 位。

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3 回答 3

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Here you have already removed an int from the stream, so you have to remove anything more. Take out the call to sc.next():

    b = sc.nextInt();
    if(b < 1 || b > 100) {
      System.out.print("That integer isn't between 1 and 100! Try again: "); 
      // sc.next(); remove this
    }

Notice how this is different from the situation with the earlier if-statement: if the user types in something that's not a number you must remove it from the stream by calling next() because nothing else would remove it otherwise. Here the call to nextInt already removes input from the stream.

于 2013-10-09T07:31:11.627 回答
0

不需要 if 语句中的 sc.next() 并使您的代码跳过用户的下一个输入。下面的代码片段可以按您的预期正常工作。

private static Scanner sc;

public static void main(String[] args) throws InterruptedException {

    sc = new Scanner(System.in);
    int a, b;
    System.out.println("Enter a number     between 1 and 100: ");

    while(true) {
        b = 0;
        if(!sc.hasNextInt()) {
            System.out.println("That isn't an integer! Try again: ");
            sc.next(); 
        }
        else {
            b = sc.nextInt();
        }
        if(b < 1 || b > 100) {
            System.out.println("That integer isn't between 1 and 100! Try again: "); 
        }
        else {
            a = b; 
            break;
        }
    }
    System.out.println("The number is: "+a+".");
    return;
}
于 2013-10-09T07:22:35.203 回答
0

尝试下面的主要它会工作

公共静态 void main(String[] args) 抛出 InterruptedException {

sc = new Scanner(System.in);
int a, b;
System.out.print("Enter a number     between 1 and 100: ");

while(true) {
  b = 0;
  if(!sc.hasNextInt()) {
    System.out.print("That isn't an integer! Try again: "); 
    sc.next();
  }
  else{
    b = sc.nextInt();
    if(b < 1 || b > 100) {
      System.out.print("That integer isn't between 1 and 100! Try again: "); 
    }
    else{
      a = b; 
      break;
    }
  }
}
System.out.print("The number is: "+a+".");

}

于 2013-10-09T07:42:59.677 回答