Re: Byte.parseByte throws NumberFormatException for value "95", radix: 16

[Janda Martin]

Thank you very much for fast reply.

  I use (byte)Integer.parseInt....

  Martin

----- Original Message -----
From: "David Holmes" <david.hol...@oracle.com>
To: "Janda Martin" <jan...@crcdata.cz>
Cc: "core-libs-dev" <core-libs-dev@openjdk.java.net>
Sent: Tuesday, July 12, 2011 2:19:25 PM GMT +01:00 Amsterdam / Berlin / Bern / 
Rome / Stockholm / Vienna
Subject: Re: Byte.parseByte throws NumberFormatException for value "95", radix: 
16

Janda Martin said the following on 07/12/11 21:32:
> I have question about parsing hex string. Java 6u25
> 
> When I call Byte.parseByte("95", 16) I get.
> 
> java.lang.NumberFormatException: Value out of range. Value:"95" Radix:16
>       at java.lang.Byte.parseByte(Byte.java:153)
> 
> 95 hex is 149 dec, Integer value '149' can be cast to valid byte value -107
> 
> Is the exception correct?

Yes. 95 hex == 149 dec which is > 127 and so is out of range for byte. 
If you want to use the bit pattern of 95 hex as a byte then you need to 
parse it as an int and do the cast yourself.

David Holmes

>   Thank you very much for answer
> 
>  Martin JANDA
>  jan...@crcdata.cz
> 
> 
> Code from Byte.java
> 
>     public static byte parseByte(String s, int radix)
>       throws NumberFormatException {
>       int i = Integer.parseInt(s, radix);
>       if (i < MIN_VALUE || i > MAX_VALUE)
>           throw new NumberFormatException(
>                 "Value out of range. Value:\"" + s + "\" Radix:" + radix);
>       return (byte)i;
>     }