«

Hi,
I am trying to decode the RLE 16 - bit compression using java.
So here I believe all the data is encoded in 16-bit. so even the face
number should be encoded in 16-bit. using java I am reading it as 2
eight bits A and B. Now I need to add these 2 bits in such a way that I
get the correct value in an 8-bit space.

any ideas?

Thank u in advance,
kind regards,
Rajesh Rapaka.

On 10 Aug 2005 10:02:30 -0700, "Rajesh.Rapaka"
<> wrote or quoted :

>number should be encoded in 16-bit. using java I am reading it as 2
>eight bits A and B. Now I need to add these 2 bits in such a way that I
>get the correct value in an 8-bit space.

I presume you mean adding modulo 256.


sum = ( a + b ) & 0xff;

--
Bush crime family lost/embezzled $3 trillion from Pentagon.
Complicit Bush-friendly media keeps mum. Rumsfeld confesses on video.
http://www.infowars.com/articles/us/...s_rumsfeld.htm

Canadian Mind Products, Roedy Green.
See http://mindprod.com/iraq.html photos of Bush's war crimes

In a previous article, look-lid said:
>On 10 Aug 2005 10:02:30 -0700, "Rajesh.Rapaka"
><> wrote or quoted :
>>number should be encoded in 16-bit. using java I am reading it as 2
>>eight bits A and B. Now I need to add these 2 bits in such a way that I
>>get the correct value in an 8-bit space.
>
>I presume you mean adding modulo 256.
>
>
>sum = ( a + b ) & 0xff;

I think he's trying to put two 8 bit values together to make one 16 bit
value, in which case he'd want

sum = a << 8 | b;



--
Paul Tomblin <> http://xcski.com/blogs/pt/
"You can get anything you want on Alice's NNTP server."

Paul Tomblin wrote:
>
> I think he's trying to put two 8 bit values together to make one 16 bit
> value, in which case he'd want
>
> sum = a << 8 | b;

Presumably not if he's stored them in bytes.

Tom Hawtin
--
Unemployed English Java programmer
http://jroller.com/page/tackline/

Paul Tomblin wrote:
>
> In a previous article, look-lid said:
> >On 10 Aug 2005 10:02:30 -0700, "Rajesh.Rapaka"
> ><> wrote or quoted :
> >>number should be encoded in 16-bit. using java I am reading it as 2
> >>eight bits A and B. Now I need to add these 2 bits in such a way that I
> >>get the correct value in an 8-bit space.
> >
> >I presume you mean adding modulo 256.
> >
> >
> >sum = ( a + b ) & 0xff;
>
> I think he's trying to put two 8 bit values together to make one 16 bit
> value, in which case he'd want
>
> sum = a << 8 | b;

At least that makes sense with his subject, assuming he got his hex to decimal
conversion wrong. (8 << 8 | is 0x88 or 136 decimal ... not 137 decimal.

--
Lee Fesperman, FFE Software, Inc. (http://www.firstsql.com)
================================================== ============
* The Ultimate DBMS is here!
* FirstSQL/J Object/Relational DBMS (http://www.firstsql.com)

Lee Fesperman <> wrote or quoted:
> Paul Tomblin wrote:
> > In a previous article, look-lid said:
> > >On 10 Aug 2005 10:02:30 -0700, "Rajesh.Rapaka"
> > ><> wrote or quoted :

> > >>number should be encoded in 16-bit. using java I am reading it as 2
> > >>eight bits A and B. Now I need to add these 2 bits in such a way that I
> > >>get the correct value in an 8-bit space.
> > >
> > >I presume you mean adding modulo 256.
> > >
> > >sum = ( a + b ) & 0xff;
> >
> > I think he's trying to put two 8 bit values together to make one 16 bit
> > value, in which case he'd want
> >
> > sum = a << 8 | b;
>
> At least that makes sense with his subject, assuming he got his hex to
> decimal conversion wrong. (8 << 8 | is 0x88 or 136 decimal ... not
> 137 decimal.

I make it 0x808.
--
__________
|im |yler http://timtyler.org/ Remove lock to reply.

yes,
I am reading them as bytes! and I am this is how i felt is the correct
way of doing it....
byte a, b;
readbytes ab();
short value = b << 8 | a;

in this particular compression technique, we get only values between
-127 and 127.

ie a and b can contain values between -127 and 127. But when i see the
value that i get i get most of them above -127.

I dont think conversion from hexadecimal to decimal is important.

any ideas?

thanks in advance,
regards,
Rajesh Rapaka.

Hi Rejesh,

Rajesh.Rapaka wrote:
> byte a, b;
> readbytes ab();
> short value = b << 8 | a;

OK, think about, what "b<<8" will do with a *byte*!

Hth,
Ingo

Ingo R. Homann wrote:
>
> Rajesh.Rapaka wrote:
>
>> byte a, b;
>> readbytes ab();
>> short value = b << 8 | a;

Doesn't even compile.

> OK, think about, what "b<<8" will do with a *byte*!

Much the same as it'd do to an int.

Think about the representation of negative ints across their 32-bits.

Tom Hawtin
--
Unemployed English Java programmer
http://jroller.com/page/tackline/

Hi,

Thomas Hawtin wrote:
> Ingo R. Homann wrote:
>>
>> Rajesh.Rapaka wrote:
>>
>>> byte a, b;
>>> readbytes ab();
>>> short value = b << 8 | a;
>
> Doesn't even compile.
>
>> OK, think about, what "b<<8" will do with a *byte*!
>
> Much the same as it'd do to an int.

Testing... compiling... testing again...

Intersting!

What's the reason, that java does not compile the first one, when it
really compiles the second one without complaining? Why does java treat
byte and int so differently in this case?:

byte b = 42;
b = b << 8; // not OK

int i = 42;
i = i << 128; // perfectly OK


> Think about the representation of negative ints across their 32-bits.

???

Ciao,
Ingo