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Java Datatypes

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  • phouthasak 6 years ago+ 0 comments

    Does anyone feel like the instructions to this challenege is kind of vague?

    182|
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  • avi89 5 years ago+ 0 comments

    A simple solution that doesn't involve the Java Math Class is to use the MIN and MAX fields on the Byte, Short, Integer, and Long wrapper classes.

     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");
                }
            }
    
    65|
    Permalink
  • hitesh_chopra 4 years ago+ 0 comments

    My Solution:

    public static void main(String[] args) {
        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)x)System.out.println("* byte");
                if(x == (short)x)System.out.println("* short");
                if(x == (int)x)System.out.println("* int");
                if(x == (long)x)System.out.println("* long");
            }
            catch(Exception e)
            {
                System.out.println(sc.next()+" can't be fitted                  anywhere.");
            }
    
        }
    }
    
    28|
    Permalink
  • CryptCode 4 years ago+ 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.");
                }
    
            }
        }
    }
    
    22|
    Permalink
  • ashishdalal_yes 3 years ago+ 0 comments

    IF INTEGER TOO LONG ,ERROR COMES FOR ANYONE:

    USE 'L' AT ENDING OF LONG.

    HERE'S HOW IT IS: if(x>=-9223372036854775808L && x<=9223372036854775807L )System.out.println("* long");

    AND UPVOTE IF IT HELPS YOU OUT.

    19|
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