如何获取Java的hasNextInt()来停止等待int而不输入字符? [英] How do I get Java's hasNextInt() to stop waiting for ints without inputting a character?

问题描述

为我提供了以下代码，并要求编写从TestList扩展的Solution类.我为此编写了一个构造函数(仅称为super)，并在下面代码的最后一行中调用了printSecond2()方法.所有其他方法都是继承的.这是代码:

I was given the following code and asked to write the Solution class extending from TestList. I wrote a constructor for it (which just called super) and the printSecond2() method invoked in the last line of the code below. All other methods are inherited. Here's the code:

public class Test3A {
    public static void main(String[] args) {
        TestList tl = new Solution();
        tl.loadList();
        ((Solution) (tl)).printSecond2();//prints every second element
    }
}

但是，该死的东西从不打印任何东西，所以我进入了TestList类(已提供)，并将println语句放在loadList()方法的每一行之后:

However, the damn thing was never printing anything out, so I went into the TestList class (which was provided) and put println statements after every single line of the loadList () method:

   public void loadList ()
   {
      if (input.hasNextInt ())//input is a Scanner object
      {
         int number = input.nextInt ();
         loadList ();
         add (number);
      }
   }

我发现我可以继续无限期地添加空格，换行符和整数，并且只有在输入字符时才最终调用add(number)方法.因此，如果我不这样做，那只是在等待更多的输入而不是继续进行而已.

I discovered that I can continue to add whitespace, newlines and integers indefinitely, and that the add(number) method is only finally called when I input a character. So if I don't do that, it just sort of hangs around waiting for more input instead of moving on.

我对此感到困惑，因为提供的示例输入/输出是:

I'm confused by this as the provided sample input/output is:

样本输入

1 2 3 4 5

样本输出

2 4

因此，自动标记器没有输入任何字符.

So there's no character being inputted by the automatic marker.

我尝试覆盖解决方案中的方法(我们无法接触其他类)，并且:

I have tried overriding the method in Solution (we can't touch the other classes) and:

1. )更改为while
2. )添加else块
3. )添加else if(！input.hasNextInt())

这些都没有改变任何东西.我不知道程序应该如何继续运行并达到调用printSecond2()的程度.

None of these changed anything. I have no idea how the program is supposed to move on and get as far as calling printSecond2().

有什么想法吗?我真的很想通过我的下一个实践测试:D

Any thoughts? I'd really like to pass my next prac test :D

推荐答案

当用户应输入数字序列时，应提供项目数或以某种方式终止输入. 1 2 3和1 2 3 4都是有效输入，因此扫描仪无法自行决定在何处结束. 可以假定，由于没有给出其他信息，所以数字序列在Windows上以EOF字符Ctrl-Z终止，在Unix上以Ctrl-D终止.

When user is supposed to enter a sequence of numbers either the number of items should be provided or the input should be terminated in some manner. 1 2 3 and 1 2 3 4 are both valid inputs so scanner can't decide where to end on its own. It can be assumed that the number sequence is terminated by EOF character Ctrl-Z on windows and Ctrl-D on unix as no other information is given.

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