Scanner.next ...为Float抛出java.util.InputMismatchException但不为Int抛出

时间:2015-03-11 20:50:44

标签: java floating-point integer java.util.scanner inputmismatchexception

为什么Java在使用Scanner.nextFloat()而不是Scanner.nextInt()时会抛出错误?

package myshit;

import java.lang.Math;
import java.util.Scanner;

public class speed2 {
    public static Scanner keyboard = new Scanner(System.in);

    public static void main(String[] args){
        float number = keyboard.nextFloat();
        System.out.print("Start");

        }
    }

输入:

2.5

输出:

Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Unknown Source)
at java.util.Scanner.next(Unknown Source)
at java.util.Scanner.nextFloat(Unknown Source)
at myshit.speed2.main(speed2.java:10)

但是只需将nextFloat切换到nextInt就不会发生错误:

package myshit;

import java.lang.Math;
import java.util.Scanner;

public class speed2 {
    public static Scanner keyboard = new Scanner(System.in);

    public static void main(String[] args){
        int number = keyboard.nextInt();
        System.out.print("Start");

        }
    }

输入:

3

输出:

Start

我做错了什么?

出现我需要输入,而不是。 似乎是因为Eclipse

2 个答案:

答案 0 :(得分:2)

你应该输入2,5而不是2.5(我认为这只发生在Netbeans,有趣的事实是它被解析为2.5)

run:
2,5
Your number is 2.5

在Netbeans中使用符号2.5。

run:
2.5

Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Scanner.java:864)
at java.util.Scanner.next(Scanner.java:1485)
at java.util.Scanner.nextFloat(Scanner.java:2345)
at test.Test.main(Test.java:25)

答案 1 :(得分:0)

我希望你不要试图从命令行给一个浮点数输入2.5f ...因为那时你得到一个错误..

除了你的代码工作正常,你可以看到你是否添加打印你输入的数字:

import java.lang.Math;
import java.util.Scanner;

public class speed2 {
    public static Scanner keyboard = new Scanner(System.in);

    public static void main(String[] args){
        System.out.print("Give me float:");
        float number = keyboard.nextFloat();
                System.out.print("You gave me the number: " + number);


        }
    }

输出

Give me float:2.5
You gave me the number: 2.5