RE: at2 swap_nibbles

 
 



-----Original Message-----
From: Andreas Fink [mailto:[email protected]]
Sent: Monday, November 25, 2002 7:44 PM
To: Oded Arbel
Cc: [email protected]
Subject: Re: at2 swap_nibbles



It matters because of this: 

return ( ( byte & 15 ) * 10 ) + ( byte >> 4 ); 

The >> operator is allowed to sign-extend its operand, so that -16 
(0xf0h) becomes -1 (0xffh). It won't do this if chars are unsigned, 
in which case 0xf0h just becomes 0x0fh. 



correct. the other option would be to OR away all the unneeded bits like this: 
the function is in any case wrong. 

Assume you got 0xF3 and you are supposed to get 0x0F 

byte & 15 = 0x03 
(byte & 15) * 10 = decimal 30 = 0x14 
( ( ( byte & 15 ) * 10 ) + ( byte >> 4 ) = 0x14 + 0x0F = 0x33. 

Is it anywhere used at all? 


 
Yes. will somebody just shoot me where I stand ? will save me lots of trouble ;-)
I shoud have said
return ( ( byte &  0x0F )  << 4 )  |  (  ( byte  & 0xF0 ) >> 4 ); 
I'll fix ?
 
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Oded Arbel
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