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Java Datatypes

  • CryptCode
     + 0 comments

    This should work without any problem and you don't have to remember range of any Datatypes. Happy Coding!

    import java.util.*;
import java.io.*;

class Solution{
    public static void main(String []argh){
        Scanner sc = new Scanner(System.in);
        int t=sc.nextInt();

        for(int i=0;i<t;i++){
            try{
                long x=sc.nextLong();
                System.out.println(x+" can be fitted in:");
                if(x>=Byte.MIN_VALUE && x<=Byte.MAX_VALUE)
                    System.out.println("* byte");
                if(x>=Short.MIN_VALUE && x<=Short.MAX_VALUE)
                    System.out.println("* short");
                if(x>=Integer.MIN_VALUE && x<=Integer.MAX_VALUE)
                    System.out.println("* int");
                if(x>=Long.MIN_VALUE && x<=Long.MAX_VALUE)
                    System.out.println("* long");
            }
            catch(Exception e){
                System.out.println(sc.next()+" can't be fitted anywhere.");
            }

        }
    }
}