| Decomposing a bit field? [message #71716] |
Wed, 21 July 2010 02:40  |
rjp23
Messages: 97
Registered: June 2010
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Member |
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Hi all,
I have some data that is contained within a 32-bit field.
I'm using some code from a colleague which seems to almost do what I
want but not quite.
An example of one of the data values is 15982561
The following code takes this and creates an array of 1s and 0s for
whether each bit is set or not.
test=intarr(32)
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.
What I need to extract is the information in the folowing form:
Bit 0: 0 or 1
Bits 1-4: 0-15
Bit 5: 0 or 1
Bits 6-8: 0-8
etc
Thanks in advance
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| Re: Decomposing a bit field? [message #71786 is a reply to message #71716] |
Wed, 21 July 2010 15:41  |
Chris[6]
Messages: 84
Registered: July 2008
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Member |
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On Jul 21, 1:25 am, Rob <rj...@le.ac.uk> wrote:
> 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?
right- 7 is (000000....111) in binary. So if you take "x and 7" it
will set all but the last three bits to zero. So it throws away all
the leading bits, and returns the last 3 bits as some number (0-7)
15 (0000.....1111) does the same thing, but with 4 bits instead of 3.
chris
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| Re: Decomposing a bit field? [message #71788 is a reply to message #71716] |
Wed, 21 July 2010 09:21 |
penteado
Messages: 866
Registered: February 2018
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Senior Member
Administrator |
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On Jul 21, 1:04 pm, Steve <f...@k.e> wrote:
>> 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?
>
> The "B" suffix to a number tells IDL that it is to be interpreted in
> octal so indeed 17B is 8+4+2+1=15 in decimal
No, B means that it is of type byte, regardless of the base used.
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| Re: Decomposing a bit field? [message #71790 is a reply to message #71716] |
Wed, 21 July 2010 09:09 |
David Fanning
Messages: 11724
Registered: August 2001
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Senior Member |
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Steve writes:
> The "B" suffix to a number tells IDL that it is to be interpreted in
> octal so indeed 17B is 8+4+2+1=15 in decimal
Well, not exactly. :-)
Octal numbers (octal numbers!?) are typically written
with a double quote in font of them:
IDL> number = "17
IDL> print, number
15
Cheers,
David
--
David Fanning, Ph.D.
Fanning Software Consulting, Inc.
Coyote's Guide to IDL Programming: http://www.dfanning.com/
Sepore ma de ni thui. ("Perhaps thou speakest truth.")
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| Re: Decomposing a bit field? [message #71791 is a reply to message #71716] |
Wed, 21 July 2010 09:04 |
Steve[5]
Messages: 10
Registered: September 2007
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Junior Member |
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>
> 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?
>
The "B" suffix to a number tells IDL that it is to be interpreted in
octal so indeed 17B is 8+4+2+1=15 in decimal
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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
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Member |
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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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| Re: Decomposing a bit field? [message #71809 is a reply to message #71716] |
Wed, 21 July 2010 03:33 |
Chris[6]
Messages: 84
Registered: July 2008
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Member |
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On Jul 20, 11:40 pm, Rob <rj...@le.ac.uk> wrote:
> Hi all,
>
> I have some data that is contained within a 32-bit field.
>
> I'm using some code from a colleague which seems to almost do what I
> want but not quite.
>
> An example of one of the data values is 15982561
>
> The following code takes this and creates an array of 1s and 0s for
> whether each bit is set or not.
>
> test=intarr(32)
> 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.
>
> What I need to extract is the information in the folowing form:
>
> Bit 0: 0 or 1
> Bits 1-4: 0-15
> Bit 5: 0 or 1
> Bits 6-8: 0-8
> etc
>
> Thanks in advance
The code you attached creates an array whose ith element is 1 if the
ith bit (i.e. the 2^i component) is 1 -- confirm with an easy number
like 5 (101 in binary). So the easiest way to combine bits 1-4 would
be something like
answer = total([1, 2, 4, 8] * test[1:4])
or, more mysteriously,
answer = ishft(number, -1) and 15B
chris
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| Re: Decomposing a bit field? [message #71810 is a reply to message #71716] |
Wed, 21 July 2010 03:31 |
greg.addr
Messages: 160
Registered: May 2007
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Senior Member |
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On 21 Jul., 11:40, Rob <rj...@le.ac.uk> wrote:
> Hi all,
>
> I have some data that is contained within a 32-bit field.
>
> I'm using some code from a colleague which seems to almost do what I
> want but not quite.
>
> An example of one of the data values is 15982561
>
> The following code takes this and creates an array of 1s and 0s for
> whether each bit is set or not.
>
> test=intarr(32)
> 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.
>
> What I need to extract is the information in the folowing form:
>
> Bit 0: 0 or 1
> Bits 1-4: 0-15
> Bit 5: 0 or 1
> Bits 6-8: 0-8
> etc
>
> Thanks in advance
The code you have is progressively shifting the bits to the right
(i.e. dividing by 2^i) and copying the resulting least significant bit
into the test array. You can achieve what you want with something like
this:
IDL> a=15982561
IDL> print,(a and ishft(1,indgen(32))) gt 0 ;decomposition like yours
(least significant on the left)
1 0 0 0 0 1 1 1 1 1 1 1 1 0 1 1 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
IDL> bit=[0,1,5,6] ;start bit for each group
IDL> n_bits=[1,4,1,3] ;no. of bits in each
IDL> print,2^n_bits-1 ;max values for each group
1 15 1 7
IDL> print,2^n_bits-1 and ishft(a,-bit) ;the result
1 0 1 7
regards,
Greg
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| Re: Decomposing a bit field? [message #71811 is a reply to message #71716] |
Wed, 21 July 2010 03:29 |
Steve[5]
Messages: 10
Registered: September 2007
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Junior Member |
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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
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