Re^2: Extracting Bit Fields (Again)

There is a rotate... I'd just forgotten about it... and this way is the simplest and clearest for what I need to do:-

$D1 = ($intbuf & 0x80) ? 1 : 0;
$D2 = ($intbuf >> 4) & 0x07; 
$D3 = ($intbuf & 0x0F);
[download]

My fogged brain can now keep struggling away at this job...

Many, many thanks for all your help, folks.

Comment on Re^2: Extracting Bit Fields (Again) Download Code

Replies are listed 'Best First'.
Re^3: Extracting Bit Fields (Again) by ikegami (Patriarch) on Oct 12, 2005 at 14:17 UTC
There is a rotate... huh? no Or are you talking about some module?	[reply]
Re^4: Extracting Bit Fields (Again) by ozboomer (Friar) on Oct 13, 2005 at 10:20 UTC
Sorry about that... my slack use of the language(!) I mean a 'right shift operator' (the '>>'). Sometimes I revert to assembler or C -style constructs/language and I use the terms interchangably... but it's really just 'rotating the bits (right)' that I was trying to remember.	[reply]
Re^5: Extracting Bit Fields (Again) by ikegami (Patriarch) on Oct 13, 2005 at 13:17 UTC
C doesn't have a rotate operator either. The rotate operator takes the bits that are shifted out of the register and shifts then in from the other end. Rotate 010...101 by 1 to to the right = 1010...10 Shift 010...101 by 1 to to the right = 0010...10	[reply]