Hi
If I have a hex value "AC23", whose 16-bit binary is "1010110000100011" and decimal is "44067". How can I break it into following groups of bits (bit-fields):
(5-bits)--(1-bit)--(5-bits)--(5-bits)
10101 -- 1 -- 00001-- 00011
21 -- 1 -- 1 -- 3
Is there any formula by which I can break a HEX value into its bits as above and then I can get those values in columns as shown in attached image?
bitf.jpg
Do you absolutely need to do this with formulas or would a VBA macro be acceptable?
For whichever answer you give me for the above question, what is the range of Hex numbers that can appear in a cell?
Also, will you need to do this for only cell A1 or for a range of cells in Column A?
or
=bin2dec(mid(text(hex2bin(left(a1,2)),rept(0,8))&text(hex2bin(right(a1,2)),rept(0,8)),1,5))=BIN2DEC(LEFT(TEXT(HEX2BIN(LEFT(A1,2)),REPT(0,8)),5))or=bin2dec(mid(text(hex2bin(left(a1,2)),rept(0,8))&text(hex2bin(right(a1,2)),rept(0,8)),6,1))=BIN2DEC(MID(TEXT(HEX2BIN(LEFT(A1,2));REPT(0,8)),6,1))=bin2dec(mid(text(hex2bin(left(a1,2)),rept(0,8))&text(hex2bin(right(a1,2)),rept(0,8)),7,5))or=bin2dec(mid(text(hex2bin(left(a1,2)),rept(0,8))&text(hex2bin(right(a1,2)),rept(0,8)),12,5))=BIN2DEC(RIGHT(TEXT(HEX2BIN(RIGHT(A1,2)),REPT(0,8)),5))
Last edited by BMV; 02-01-2020 at 12:37 PM.
Thanks BMV
Can you explain a bit about the first of the two options that you have provided?
Assuming the value in cell A1 is always 4 hex-digits long, put these formulas in the indicated cells (and copy down if necessary)...
B1: =BIN2DEC(LEFT(HEX2BIN(LEFT(A1,2),8),5))
C1: =0+MID(HEX2BIN(LEFT(A1,2),8),6,1)
D1: =BIN2DEC(MID(HEX2BIN(LEFT(A1,2),8)&HEX2BIN(RIGHT(A1,2),8),7,5))
E1: =BIN2DEC(RIGHT(HEX2BIN(RIGHT(A1,2),8),5))
Last edited by Rick Rothstein; 02-02-2020 at 02:33 AM.
oh. I forgot about 2nd parameter.
Rick Rothsteini, I thin it's typo =0+MID(HEX2BIN(LEFT(A1,2),8),6,1)
Last edited by BMV; 02-02-2020 at 01:13 AM.
Yes, it was a typo... I have corrected it in Message #4 so it now reads correctly. Thank you for catching that.Originally Posted by BMV
linko, Use Rick Rothstein formulas. However
My first variant the same as
D1: =BIN2DEC(MID(HEX2BIN(LEFT(A1,2),8)&HEX2BIN(RIGHT(A1,2),8),7,5)) and only last numbers 7 and 5 should be change according you rules (5-bits)--(1-bit)--(5-bits)--(5-bits). So it looks like universal.
The second option is only short way.
Can you tell any specific reason why you suggesting to use Rick Rothstein's formulas?
Last edited by AliGW; 02-02-2020 at 05:49 AM. Reason: Please don't quote unnecessarily!
not always shorter formula better from performance point of view, but if it possible to use one standard function in stand of two or more it's best way.
Rick's fcormulae are more compact, therefore easier to maintain going forward.
Ali
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I didn't say "shorter" - I said "more compact". Not the same thing.
AliGW. it was additional comment only.
Just came here on the back of the reverse question I answered. Perhaps more compact formulas with minimal calculations are:
WBD=BITAND(HEX2DEC(A1),31*2048)/2048 =BITAND(HEX2DEC(A1),1024)/1024 =BITAND(HEX2DEC(A1),31*32)/32 =BITAND(HEX2DEC(A1),31)
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Sorry. Just amusing myself now. Pure bit manipulation (old programmers and all that ...):
:D=BITRSHIFT(BITAND(HEX2DEC(A1),BITLSHIFT(31,11)),11) =BITRSHIFT(BITAND(HEX2DEC(A1),BITLSHIFT(1,10)),10) =BITRSHIFT(BITAND(HEX2DEC(A1),BITLSHIFT(1,10)),10) =BITAND(HEX2DEC(A1),31)
WBD
My version of Excel (2010) does not have those first two functions.
BITAND and BITRSHIFT only from Excel 2013
Ah. OK.
BITLSHIFT(number,shift) = number*2^shift
BITRSHIFT(number,shift) = INT(number/2^shift)
so...
is=BITRSHIFT(BITAND(HEX2DEC(A1),BITLSHIFT(31,11)),11)
As I said, I was just messing around with bits and bobs ...=BITAND(HEX2DEC(A1),31*2^11)/2^11 or =BITAND(HEX2DEC(A1),63488)/2048
:D
WBD
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