Counting bits

[RussMartin]

Does anyone know a quick and easy (or sneaky) way to count bits?

I want to look at either the top or bottom 4 bits of a word and determine whether an odd or even number of bits are high (or low; doesn't matter).

Right now, I'm using LOOKDOWN and it works just fine; but I keep thinking there's a more elegant way to do it.

[Darrel Taylor]

Hi Russ,

Not particularly "Sneaky" ... but it's "Quick".

Counter VAR BYTE BANK0
ODD VAR Counter.0
TestWord VAR WORD BANK0

ASM
clrf _Counter
btfsc _TestWord,0
incf _Counter, F
btfsc _TestWord,1
incf _Counter, F
btfsc _TestWord,2
incf _Counter, F
btfsc _TestWord,3
incf _Counter, F
ENDASM

If ODD THEN
; ODD number of bits in the TestWord<3:0>
ENDIF

[Acetronics2]

Hi, Russ

first, shift right your word 12 times if > 4095 ...

mask the result to keep the 4 lower bits,

then 6 possibilities remain for an even count of the bits ...

( call the result "ABCD" ... if even, ABCD = Ch,Ah,9h,6h,5h or 3h )

A SELECT CASE will do it.

mask your word to keep the 4 lower bits

Then 6 possibilities ...

The SELECT CASE ...

it's one way ...in PbP !!! ( lol )


or use "NCD" function in a loop ... which looks furiously to what Darrel Wrote !!!


Alain

[RussMartin]

Thanks to both of you.

I'm emulating a 16-bit twisted ring counter--actually a very twisted counter. If an even number of bits in the top four are high, the whole works is shifted left one place and the lowest bit is set to 1; if an odd number of bits are high, again a one-place shift left, but the default 0 shifted in is retained.

Just now, the target device is a 16F87xA.