Adding bits in variables

Moskog

Hello,
Now I have this little problem, it is a part of a simple validation routine in a transmitter-reciever environment! I want the transmitter to send a number (checksum) along with·one or·more
variables. The checksum is all the bits in all the variables added together,
like this:· something = 13······binary 1101······ checksum = 3.

This is an example on how to find the checksum of the variable Something:

something VAR Nib
checksum VAR Nib

checksum = 0
checksum = something.BIT0 + something.BIT1 + something.BIT2 + something.BIT3


But this is not a good programming-solution if I have·word-sized variables or even several of them.
Being looking after a better way but can't seem to find anything in my books on adding bits in variables.
Anyone there with a good idea?

KjellO

Peter Verkaik

Use a for loop to divide something by 2, before division add bit0 to a counter.

something var word
count var byte
index var byte
count = 0
for index=0 to 15
· count = count + something.0
· something = something /2· 'or use shiftright by 1 position
next

regards peter

Moskog

Well.. I guess Count was a bad name of a variable here, but replacing it with a different name ended up with the error message Expected a variable modifier on "something.0".
Guess I can't say "something.index" either, would probably make same error message. (I'm using PBasic 2.5)

KjellO

Mike Green

Use "something.bit0"

Moskog

Yes, now it works, but Something disappeard into zero, I have to remember to save it before prosessing!

Thank you so much and have a nice weekend!

KjellO

Tracy Allen

x = something  ' a word variable
x = (x & $5555) + (x>>1 & $5555)
x = (x & $3333) + (x>>2 & $3333)
x = (x & $0f0f) + (x>>4 & $0f0f)
x = (x & $00ff) + (x>>8 & $00ff)
' x is now = number of ones in somthing

Which can be simplified somewhat using methods shown in Chapter 5 of hacker's delight which is all on the subject of counting bits, including tricks for larger arrays.

Another method is unique to the PBASIC interpreter, using the DCD and NCD operators. The execution time of this is proportional to the number of ones in the word:

x = something
cs = 0
DO while x
  x = x - DCD (NCD x -1)
  cs = cs + 1
LOOP
' cs is a count of the number of ones in something

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Tracy Allen
www.emesystems.com

Moskog

Yess! I'm on my knees in the dust, thats an awesome solution! Thank's so much!

KjellO