为什么input.nextint方法有一个\ n剩余呢?

本文介绍了为什么input.nextint方法有一个\ n剩余呢?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

在回答这个问题时，我不明白为什么input.nextInt的剩余字符是换行符.

In the answer to this question I don't understand why the input.nextInt has a newline character as a leftover.

扫描仪在使用next()，nextInt()或其他nextFoo()吗?

推荐答案

答案很简单:

Scanner#nextInt()读取下一个整数，但不读取用户按下以提交整数的换行符(ENTER).

Scanner#nextInt() reads the next integer but not the newline (ENTER) the user presses to submit the integer.

要了解这一点，您必须知道您键入的所有内容都会写入一个缓冲区，Scanner方法会尝试从该缓冲区中读取.

To understand this you must know that everything you type will be written to a buffer, from which the Scanner methods try to read from.

关于以下示例:

System.out.println("Enter a number: "); int i = scanner.nextInt(); System.out.println("Enter a line: "); String l = scanner.nextLine();

会发生什么:

• 用户看到Enter a number:，然后按17和ENTER.
• 扫描仪将读取17，但保留ENTER的换行符.
• 程序将写入Enter a line:，但是扫描程序将读取ENTER的换行符，而不是等待输入.

• The user sees Enter a number: and will press 17 and ENTER.
• The scanner will read the 17 but leave the newline of the ENTER.
• The program will write Enter a line: but the scanner will read the newline of the ENTER instead of waiting for input.