CodeGym/Help with Java Tasks/java.lang.NumberFormatException: For input string: "exit"...

Level 20

11.03.2021
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java.lang.NumberFormatException: For input string: "exit"

Question about the task

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output: java.lang.NumberFormatException: For input string: "exit"

1. Read data from the console until the word "exit" is entered.
2. For each value (except "exit"), call the print method. If the value:
2.1. contains a period ("."), then call the print method for Double;
2.2. is greater than zero, but less than 128, then call the print method for short;
2.3. less than or equal to zero or greater than or equal to 128, then call the print method for Integer;
2.4. otherwise, call the print method for String.

Requirements:

The program must read data from the keyboard.
The program should stop reading data from the keyboard after "exit" is entered.
If the entered string contains a period (".") and can be correctly converted to a Double, the print(Double value) method should be called.
If the entered string can be correctly converted to a short and the resulting number is greater than 0, but less than 128, the print(short value) method should be called.
If the entered string can be correctly converted to an Integer and the resulting number is less than or equal to 0 or greater than or equal to 128, the print(Integer value) method should be called.
In all other cases, the print(String value) method should be called.

package com.codegym.task.task15.task1519; import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.IOException; /* Different methods for different types */ public class Solution { public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); String data; while (true) { data = reader.readLine(); if (data.equals("exit")) { break; }} try { if (Short.parseShort(data) > 0 && Short.parseShort(data) < 128) { short a = Short.parseShort(data); print(a); } else if (Integer.parseInt(data) <= 0 || Integer.parseInt(data) >= 128) { int b = Integer.parseInt(data); print(b); } else if (data.contains(".")) { double c = Double.parseDouble(data); print(c); } else { print((String)data); } } catch (Exception e) { System.out.println(e); } } public static void print(Double value) { System.out.println("This is a Double. Value: " + value); } public static void print(String value) { System.out.println("This is a String. Value: " + value); } public static void print(short value) { System.out.println("This is a short. Value: " + value); } public static void print(Integer value) { System.out.println("This is an Integer. Value: " + value); } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

11 March 2021, 21:53

#1 you need to format your code properly When formatted properly this is what it looks like:

public static void main(String[] args) throws IOException {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    String data;
    while (true) {
        data = reader.readLine();
        if (data.equals("exit")) {
            break;
        }
    }

    try {
        if (Short.parseShort(data) > 0 && Short.parseShort(data) < 128) {
            short a = Short.parseShort(data);
            print(a);
        } else if (Integer.parseInt(data) <= 0 || Integer.parseInt(data) >= 128) {
            int b = Integer.parseInt(data);
            print(b);
        } else if (data.contains(".")) {
            double c = Double.parseDouble(data);
            print(c);
        } else {
            print((String) data);
        }
    } catch (Exception e) {
        System.out.println(e);
    }
}

Now, the while loop reads data until "exit" is entered. At that point the loop breaks and the next part of the code tries to take "exit" and parse it to a number, which is resulting in the Exception as "exit" is clearly not a number.

Libby

Level 20 , United Kingdom

11 March 2021, 22:17

so do I need to put the try/catch block in the while loop?

Libby

Level 20 , United Kingdom

11 March 2021, 22:20

It works only for the short and integer?

Guadalupe Gagnon

Level 37 , Tampa, United States

12 March 2021, 02:14

What you are experiencing is the process of testing your code. In this case results are expected because the code starts by first trying to convert the input String into a Short and if you try parsing a String such as "123.4", or a String not a number, into a short you will get a number format exception.

Libby

Level 20 , United Kingdom

12 March 2021, 09:37

I changed around the order and it works for everything except for String: This is a Double. Value: 1.1 This is a short. Value: 100 This is an Integer. Value: -1 java.lang.NumberFormatException: For input string: "test" (I also tried with removing the last else statement)

Nouser

Level 36 , Germany

12 March 2021, 10:44

What I did... 1. parsing the input for a double a) it is not a double, then it has to be a string, print(string) b) it's a double - if it contains a . print(double) - if it is >0 && <128 print((short) input) - else print((int) input) or you first parse for int - if >0 && <128 print(short) - else print(int) no int? parse for double - yes, print(double) - no print(string) but here you need to parse two times and CG already gave you the hint with the dot

Libby

Level 20 , United Kingdom

12 March 2021, 14:33

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    String data;
    while (true) {
        data = reader.readLine();
        if (data.equals("exit")) {
            break;
        }
        if (data.contains(".")) {
            try {
                double a = Double.parseDouble(data);
                print(a);
            } catch (Exception e) {
                System.out.println(e);
            }
        } else {
            try {
                int b = Integer.parseInt(data);
                if (b > 0 && b < 128) {
                    short c = Short.parseShort(data);
                    print(c);
                } else {
                     print(b);
                }
            } catch (Exception e) {
                    System.out.println(e);
            }
        }
    }
}

The same output as above. please tell me were I am wrong.

Nouser

Level 36 , Germany

12 March 2021, 15:11

solution

Pseudocode how I'd setup my first suggestion. Do you need pseudocode for the second approach as well?

while
    input
    try
        Double.parseDouble()
        //if we get here, it is a number - maybe int, short, double, float
        if (theNumber contains a dot)
             //do double related stuff
        else if (>0 && <128)
            //short
        else
             //has to be treated as an int

    catch
        //if the input is a string, parseDouble will throw an exception and we end here
        //do string related stuff

Guadalupe Gagnon

Level 37 , Tampa, United States

12 March 2021, 15:19

@Libby You are catching the exception and printing it. Notice in Nouser's example how if the parse to a number fails then in the catch block he treats the input as if it were a String and not a number. This is what you are missing in your code.

Nouser

Level 36 , Germany

12 March 2021, 15:20

To make your code work you probably need to change the handling inside the catch block, don't output the exception but the string and you could code that a lot easier, staring with that

short c = Short.parseShort(data);
print(c);

print ((short) b)

Libby

Level 20 , United Kingdom

12 March 2021, 16:05

@Nouser & Guadalupe Gagnon; Thank you so much to both of you for your help. now I completely got it and I fully understand what I did. your explanations are priceless!!