HasNextInt()无限循环 [英] HasNextInt() Infinite loop
问题描述
//input: multiple integers with spaces inbetween
Scanner sc = new Scanner(System.in);
while(sc.hasNextInt())
{
//add number to list
}
sc.hasNextInt()
正在等待integer
.仅当您输入non-integer
字符时,该字符才会出现.
sc.hasNextInt()
is waiting for an integer
. It only breaks out if you input a non-integer
character.
不久前我在这里看到了一个解决方案,但是我找不到了.
I saw a solution here before not too long ago but i cant find it anymore.
解决方案(如果您问我最好)是使用两台扫描仪.我似乎无法弄清楚它是如何使用两个扫描仪来解决此问题的.
The solution (was the best if you ask me) was using two scanners. I cant seem to figure out how it used two scanners to go around this problem.
sc.NextLine()
可能是
用户可以输入多个整数,其数量未知.例如:3 4 5 1.在它们之间有一个空格.我要做的就是在使用两个扫描仪的同时读取整数并将其放在列表中.
A user can input multiple integers the amount is unknown. Ex: 3 4 5 1. There is a space inbetween them. All i want to do is read the integers and put it in a list while using two scanners.
推荐答案
根据您的评论
用户可以输入多个整数,其数量未知.例如:3 4 5 1.在它们之间有一个空格.我要做的就是在使用两个扫描仪的同时读取整数并将其放在列表中.
A user can input multiple integers the amount is unknown. Ex: 3 4 5 1. There is a space inbetween them. All i want to do is read the integers and put it in a list while using two scanners.
您可能正在寻找:
- 扫描程序,它将从用户读取行(并且可以在需要时等待下一行)
- 另一台扫描仪,它将处理将从行中分离出来的每个数字.
- scanner which will read line from user (and can wait for next line if needed)
- another scanner which will handle splitting each number from line.
因此您的代码可能类似于:
So your code can look something like:
List<Integer> list = new ArrayList<>();
Scanner sc = new Scanner(System.in);
System.out.print("give me some numbers: ");
String numbersInLine = sc.nextLine();//I assume that line is in form: 1 23 45
Scanner scLine = new Scanner(numbersInLine);//separate scanner for handling line
while(scLine.hasNextInt()){
list.add(scLine.nextInt());
}
System.out.println(list);
这篇关于HasNextInt()无限循环的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!