CODEWITHSUNDEEP
java
Abhilash
Abhilash

Reputation: 11

How can I solve this program in another way? I want to know the type of input given by user

This is the program that I have written & I want to solve this in some other way. I don't want to use the try-catch statement.

I have researched about other ways but it was not clearly explained.

import java.io.*;
import java.util.*;
public class TPJava {
    public static void main(String args[]) throws Exception {

        Scanner scan = new Scanner(System.in);
        try {

            int int_var = scan.nextInt();
            System.out.println("It is an Integer.");
        }
        catch (InputMismatchException e)
        {
            try
            {
                String str_var = scan.next();
                System.out.println("It is a String");
            }
            catch (InputMismatchException ie)
            {

                try
                {
                    Float f = scan.nextFloat();
                    System.out.println("It is Float");

                }
                catch (InputMismatchException ime)
                {
                    System.out.println("Wrong Input.");
                }

            }

        }
    }
}

Upvotes: 0

Views: 51

Answers (1)

FlexXx
FlexXx

Reputation: 81

if (XXX instanceof int)
{
   System.out.println("It is an integer.");
}
else if (XXX instanceof String)
{
   System.out.println("It is a string.");
}
....

(and so on)

EDIT: found a solution that should work for exactly your code.

Scanner scan= new Scanner(System.in);
if(scan.hasNextInt())
{
   System.out.println("It is an integer.");
}
else if (scan.hasNextFloat())
{
   System.out.println("It is a Float.");
}
....
else
{
   System.out.println("It is a String.");
}

this checks if the next variable can be interpreted as a integer/float/ etc..... i think there is none for "String" so i just used the "else" statement there

BUT: you have to user scan.next() or nextInt or something like that afterwards, else it will continuesly use the same input over and over again (i assume)

Upvotes: 1

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