为什么 input.nextint 方法有一个 作为剩余部分? [英] Why does input.nextint method have a as a leftover?

问题描述

在这个问题的答案中，我不明白为什么 input.nextInt 有一个换行符作为剩余.

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

扫描仪正在跳过 nextLine() 之后使用 next()、nextInt() 或其他 nextFoo()?

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.

要理解这一点，您必须知道您键入的所有内容都将写入缓冲区，Scanner 方法尝试从中读取.

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.

看下面的例子:

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: 并按下 17 和 ENTER.
• 扫描器将读取 17 但保留 ENTER 的换行符.
• 程序会写Enter a line:，但扫描器会读取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.

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