Java 使用 Scanner 在一行上接受多个整数

Accepting multiple integers on a single line using Scanner

提问by chopper draw lion4

The user needs to enter a certain number of integers. Rather them enter an integer at a time, I want to make it so they can enter multiple integers on a single line, then I want those integers to be converted in an array. For example, if the user enters: 56 83 12 99then I want an array to be created that is {56, 83, 12, 99}

用户需要输入一定数量的整数。相反，他们一次输入一个整数，我想让他们可以在一行中输入多个整数，然后我希望将这些整数转换为数组。例如，如果用户输入：56 83 12 99那么我想要创建一个数组，它是{56, 83, 12, 99}

In other languages like Python or Ruby I would use a .split(" ")method to achieve this. No such thing exist in Java to my knowledge. Any advice on how to accept user input and create an array based on that, all on a single line?

在 Python 或 Ruby 等其他语言中，我会使用一种.split(" ")方法来实现这一点。据我所知，Java 中不存在这样的东西。关于如何接受用户输入并基于该输入创建数组的任何建议，所有这些都在一行中？

回答by Makoto

String#splitexists, but you have to do more work, since you're only getting strings back.

String#split存在，但你必须做更多的工作，因为你只得到字符串。

Once you've got the split, convert each element into an int, and place it into the desired array.

分割后，将每个元素转换为int，并将其放入所需的数组中。

final String intLine = input.nextLine();
final String[] splitIntLine = intLine.split(" ");
final int[] arr = new int[splitIntLine.length];
for(int i = 0; i < splitIntLine.length; i++) {
    arr[i] = Integer.parseInt(splitIntLine[i]);
}
System.out.println(Arrays.toString(arr));  // prints contents of your array

回答by mauris

Using the Scanner.nextInt()method would do the trick:

使用该Scanner.nextInt()方法可以解决问题：

Input:

输入：

56 83 12 99

56 83 12 99

Code:

代码：

import java.util.Scanner;

class Example
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int[] numbers = new int[4];
        for(int i = 0; i < 4; ++i) {
            numbers[i] = sc.nextInt();
        }
    }
}

At @user1803551's request on how Scanner.hasNext()can achieve this:

在@user1803551 关于如何Scanner.hasNext()实现这一点的请求：

import java.util.*;

class Example2
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        ArrayList<Integer> numbers = new ArrayList<Integer>();
        while (sc.hasNextInt()) { // this loop breaks there is no more int input.
            numbers.add(sc.nextInt());
        }
    }
}

回答by user1803551

The answer by Makoto does what you want using Scanner#nextLineand String#split. The answer by mauris uses Scanner#nextIntand works if you are willing to change your input requirement such that the last entry is not an integer. I would like to show how to get Scanner#nextLineto work with the exact input condition you gave. Albeit not as practical, it does have educational value.

Makoto 的答案可以使用Scanner#nextLine和做您想做的事情String#split。Scanner#nextInt如果您愿意更改输入要求，使最后一个条目不是整数，那么mauris 的答案就会使用并起作用。我想展示如何Scanner#nextLine使用您提供的确切输入条件。虽然不那么实用，但它确实具有教育价值。

public static void main(String[] args) {

    // Preparation
    List<Integer> numbers = new ArrayList<>();
    Scanner scanner = new Scanner(System.in);
    System.out.println("Enter numbers:");

    // Get the input
    while (scanner.hasNextInt())
        numbers.add(scanner.nextInt());

    // Convert the list to an array and print it
    Integer[] input = numbers.toArray(new Integer[0]);
    System.out.println(Arrays.toString(input));
}

When giving the input 10 11 12upon first prompt, the program stores them (Scannerhas a privatebuffer), but then keeps asking for more input. This might be confusing since we give 3 integers which loop through hasNextand expect that when the 4th call is made there will be no integer and the loop will break.

10 11 12在第一次提示时给出输入时，程序存储它们（Scanner有一个private缓冲区），但随后不断要求更多输入。这可能会令人困惑，因为我们给出了 3 个循环通过的整数，hasNext并期望在进行第 4 次调用时将没有整数并且循环将中断。

To understand it we need to look at the documentation:

要理解它，我们需要查看文档：

Both hasNextand nextmethods [and their primitive-type companion methods] may blockwaiting for further input. Whether a hasNextmethod blocks has no connection to whether or not its associated nextmethod will block.

既hasNext和next方法[及其基本类型companion方法]可以阻挡等待进一步的输入。一个hasNext方法是否阻塞与其关联的next方法是否会阻塞没有关系。

(emphasis mine) and hasNextInt

（强调我的）和 hasNextInt

Returns: trueif and only if this scanner's next token is a valid int value

返回：true当且仅当此扫描器的下一个标记是有效的 int 值

What happens is that we initialized scannerwith an InputStream, which is a continuousstream of data. On the 4th call to hasNextInt, the scanner "does not know" if there is a next int or not because the stream is still open and data is expected to come. To conclude from the documentation, we can say that hasNextInt

发生的事情是我们scanner用一个初始化InputStream，它是一个连续的数据流。在第 4 次调用 时hasNextInt，扫描器“不知道”是否有下一个 int 值，因为流仍处于打开状态并且预计会有数据到来。从文档中得出结论，我们可以说hasNextInt

Returns trueif this scanner's next token is a valid int value, returns falseif it is not a valid int, and blocks if it does not know what the next token is.

返回true此扫描器的下一个标记是否是有效的 int 值，false如果它不是有效的 int 则返回，如果不知道下一个标记是什么则阻塞。

So what we need to do is close the stream after we got the input:

所以我们需要做的是在我们得到输入后关闭流：

// Get the input
numbers.add(scanner.nextInt());
System.in.close();
while (scanner.hasNextInt())
    numbers.add(scanner.nextInt());

This time we ask for the input, get all of it, close the stream to inform scannerthat hasNextIntdoes not need to wait for more input, and store it through iteration. The only problem here is that we closed System.in, but if we don't need more input it's fine.

这一次，我们要求的输入，获得这一切，靠近流，通知scanner说hasNextInt并不需要等待更多的投入，并将其存储通过迭代。这里唯一的问题是我们关闭了System.in，但如果我们不需要更多输入，那就没问题了。