Java Hex Number and Conditionals

I am new to Java so I have a problem with hex numbers usage in conditionals and its sizes. There is some issue about Scanner class too. I have searched java documents for primitive data types.

The program is about; take a number and look in which data type it could be written and print the appropriate data type.

import java.util.*;
import java.io.*;
import java.lang.Integer;


public class Solution{
   public static void main(String []argh)
    {
      Scanner sc = new Scanner(System.in);
      long 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>=0x81 || x<=0x7f)System.out.println("* byte");
            if(x<=0x7fff || x>=0x8001)System.out.println("* short");
            if(x<=0x7fffffff || x>=0x80000001)System.out.println("* int");

            //if(x<= (0x7fffffffffffffff)|| x>= 
                       //(0x8000000000000001))System.out.println("*long");
            System.out.printf("\n\n%x\n\n",x);
            System.out.println();
        }
            catch(Exception e)
        {
            System.out.println(sc.next()+" can't be fitted anywhere.");
        }

      }
   }
 }

Questions:

1. I get an error message for commented "if()" conditional "too large int :0x7fffffffffffffff " why is that so?

2. I run the program by commenting some "if()" and give it "10" to search possible data types but this time it prints just only "int" datatype.

3. If I give it "0x7b" as input it carries out to "catch()" section. Why?

Could you please explain it ? Thanks.

I get an error message for commented "if()" conditional "too large int :0x7fffffffffffffff " why is that so?

Because 0x7fffffffffffffff is an int literal. 7fffffffffffffff can't fit into an int. It can fit into a long though. To make it a long literal, you have to add l or L at the end:

if(x<= (0x7fffffffffffffffL)|| x>=
    (0x8000000000000001L))System.out.println("*long");

But this if statement is redundant, because x is already of type long, so by definition, its value must fit in long.

Actually, using the MAX_VALUE and MIN_VALUE constants of each type is less error-prone to write than numeric literals:

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");
System.out.println("*long");

I run the program by commenting some "if()" and give it "10" to search possible data types but this time it prints just only "int" datatype.

I cannot reproduce this. Which if statements did you comment out?

If I give it "0x7b" as input it carries out to "catch()" section. Why?

Scanner.nextLong does not recognise hex numbers, unfortunately. You can read the number as a string, check its prefix, if it's 0x, then strip off the first two characters, and parse it as a hex number.