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Re: Decomposing a bit field? [message #71807 is a reply to message #71716]

Wed, 21 July 2010 04:25

rjp23
Messages: 97
Registered: June 2010

Member

On Jul 21, 11:29 am, Steve <f...@k.e> wrote:
> Rob wrote:
>> for i=0, 31 do begin test[i]=(ISHFT((15982561),-i) AND 1B)
>
>> However, for example bits 1-4 need to be combined (and contain data
>> from 0-15). As I don't quite understand what the code above is doing
>> I'm not sure how to (or if it can be) modified to not just operate on
>> the individual bits but on groups of bits.
>
> ISHFT - moves bits right or left so that the bits you want are at the
> end of the value. AND selects the bits that you want.
>
>
>
>> What I need to extract is the information in the folowing form:
>
>> Bit 0: 0 or 1
>
> bit0= ishft(15982561,0) and 1B
>
>> Bits 1-4: 0-15
>
> bits1to4 = ishft(15982561,-1) and 17B
>
>> Bit 5: 0 or 1
>
> bit5 = ishft(15982561,-5) and 1B
>
>> Bits 6-8: 0-8
>
> bits6to8 = ishft(15982561,-6) and 7B
>
>> etc
>
>> Thanks in advance
>
>

Thanks everyone, that makes much more sense now.

I just wanted to confirm where the 7B comes from above.

I assume it's because I'm checking 3 bits so it's 1+2+4. If I had a
group of 4 bits it's be 1+2+4+8 (15 like in Chris's example below),
right?

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[Message index]

		Decomposing a bit field? By: rjp23 on Wed, 21 July 2010 02:40
		Re: Decomposing a bit field? By: Chris[6] on Wed, 21 July 2010 15:41
		Re: Decomposing a bit field? By: penteado on Wed, 21 July 2010 09:21
		Re: Decomposing a bit field? By: David Fanning on Wed, 21 July 2010 09:09
		Re: Decomposing a bit field? By: Steve[5] on Wed, 21 July 2010 09:04
		Re: Decomposing a bit field? By: rjp23 on Wed, 21 July 2010 04:25
		Re: Decomposing a bit field? By: Chris[6] on Wed, 21 July 2010 03:33
		Re: Decomposing a bit field? By: greg.addr on Wed, 21 July 2010 03:31
		Re: Decomposing a bit field? By: Steve[5] on Wed, 21 July 2010 03:29

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