please review- NumberFormatException in Conversation of String to Hexadecimal

[Babu Singh]

  public class StringToHexaDecimal {

    public static void main(String[] args) {
 try 
  {  
    // attempt convert to hex 
    String s = "0cv"; 
    int i = Integer.parseInt( s,16 ) ; 
    System.out.print ( " Hex Value of: " + s + " = " + i ) ; 
  }
catch(NullPointerException npe){
System.out.println(npe);
}
catch(NumberFormatException nfe){
System.out.println(nfe);
}
}
}  

Ouput:
java.lang.NumberFormatException: For input string: "0cv"

[Omar Al Kababji]

I think you need to learn a bit about Hex numbers

0 1 2 3 4 5 6 7 8 9 A B C D E F

there is no V in a hex number, so you get the exception.

[Babu Singh]

Dear Sir,

thanks for your reply.

when i give input String s = "12345678";
Output is: Hex Value of: 12345678 = 305419896

when i give input String s = "123456789";
Output is: java.lang.NumberFormaatException: For input string: "123456789"

[Omar Al Kababji]

ok this is another issue now your problem is that

123456789 Hex = 4886718345 in Decimal and this value is too big to fit in an integer, so use a long instead of int

Long.parseLong();

[Babu Singh]

Dear Sir,

thank you to resolve my queries.

[Omar Al Kababji]

No problem