Scanner.nextInt(x) throws PatternSyntaxException

Question

i know im so nooby for asking but please i want to pass my midterm

one of the many questions i have a problem with is ""Write an if-else statement that assigns 20 to the variable y if the variable x is greater than 100. Otherwise, it should assign 0 to the variable y."

mine looks like this

import java.util.Scanner;

public class ifelseprac1 {

    public static void main(String[] args) {
        int y;
        int x=0;

        Scanner keyboard = new Scanner(System.in);

        System.out.print("Enter a number for x: ");
        x=keyboard.nextInt(x);

        if (x>100)
            y=20;           
        else 
            y=0;
    }
}

the error code im getting is

Configuration: ifelseprac1 - JDK version 1.8.0_25

Enter a number for x: Exception in thread "main" java.util.regex.PatternSyntaxException: Unclosed character class near index 525

I fail to see how the code you have pasted so far can throw this exception, since you don't use regexes at all — fge, Nov 05 '14 at 05:08
thanks for the speedy answers.. this is incredible.. would you guys happen to have any tips on how to complete this problem? — Granit Lahu, Nov 05 '14 at 05:10
@fge: apparently `nextInt` does internally. That's the beauty of stack traces. — Thilo, Nov 05 '14 at 05:16
@Thilo argh... Good point, I didn't think `Scanner` would resort to such a mechanism! — fge, Nov 05 '14 at 05:24

score 4 · Accepted Answer · edited Nov 05 '14 at 05:26

4

Instead of doing:

x = keyboard.nextInt(x);

Do:

x = keyboard.nextInt();

If you pass a parameter to nextInt, it's interpreted as the radix of the entered integer.

I'm guessing that, for certain radices, the regex that Java generates to extract the integer is incorrect, but would need to debug the JDK to be sure.

edited Nov 05 '14 at 05:26

Unihedron

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answered Nov 05 '14 at 05:09

Robby Cornelissen

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WOW it worked!!!!thank you so much! and btw wouldnt a string need to be inserted between the parenthesis where the x was? – Granit Lahu Nov 05 '14 at 05:13
There's no `nextInt` method that takes a string argument. – Robby Cornelissen Nov 05 '14 at 05:20

score 0 · Answer 2 · edited Nov 05 '14 at 05:25

0

It should be like this:

x = keyboard.nextInt();

edited Nov 05 '14 at 05:25

Unihedron

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answered Nov 05 '14 at 05:11

Kshitij Kulshrestha

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Scanner.nextInt(x) throws PatternSyntaxException

mine looks like this

2 Answers2