Java: how to read an input int

Question

So, I was looking for an efficient way, using Java's standard packages, to read an input integer... For example, I came across the class "Scanner", but I found two main difficulties:

if I don't insert an int, I'm not actually able to solve the exception;
this class works with tokens, but my aim is to load the string in its full length.

This is an example of execution I would like to realize:

Integer: eight
Input error - Invalid value for an int.
Reinsert: 8 secondtoken
Input error - Invalid value for an int.
Reinsert: 8
8 + 7 = 15

And this is the (incorrect) code I tried to implement:

import java.util.Scanner;
import java.util.InputMismatchException;

class ReadInt{
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        boolean check;
        int i = 0;
        System.out.print("Integer: ");
        do{
            check = true;
            try{
                i = in.nextInt();
            } catch (InputMismatchException e){
                System.err.println("Input error - Invalid value for an int.");
                System.out.print("Reinsert: ");
                check = false;
            }
        } while (!check);
        System.out.print(i + " + 7 = " + (i+7));
    }
}

Can you please elaborate on what you mean by if I don't insert an int, I'm not actually able to solve the exception? — Jimmy Thompson
– Jimmy Thompson, Commented Jan 10, 2013 at 14:00

xagyg · Accepted Answer · 2013-01-10 14:01:27Z

1

Use a BufferedReader. Check NumberFormatException. Otherwise very similar to what you have. Like so ...

import java.io.*;

public class ReadInt{
    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        boolean check;
        int i = 0;
        System.out.print("Integer: ");
        do{
            check = true;
            try{
                i = Integer.parseInt(in.readLine());
            } catch (NumberFormatException e){
                System.err.println("Input error - Invalid value for an int.");
                System.out.print("Reinsert: ");
                check = false;
            }
        } while (!check);
        System.out.print(i + " + 7 = " + (i+7));
    }
}

answered Jan 10, 2013 at 14:01

xagyg

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Comments

tckmn · Accepted Answer · 2013-01-10 13:58:15Z

1

To use with tokens:

int i = Integer.parseInt(in.next());

Then you could do:

int i;
while (true) {
    System.out.print("Enter a number: ");
    try {
        i = Integer.parseInt(in.next());
        break;
    } catch (NumberFormatException e) {
        System.out.println("Not a valid number");
    }
}
//do stuff with i

That above code works with tokens.

answered Jan 10, 2013 at 13:58

tckmn

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1 Comment

Marco Over a year ago

It works fine but I don't want to consider tokens.... For example, if I set as input "hello 8 testtest" I don't want to have i = 8, but instead an error, because the whole input is a string in this case

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