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?


Andreas Fink
Fink Consulting GmbH

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