Another way to left shift a 32 bit data?

[richard]

you can set or clr any bit in an array
dis var byte [4]

dis.0[x] = 1 ; will set bit x in array dis where x=0 to 31

or make a user command something like this (untested)
USERCOMMAND "SETBIT" ; BIT,VALUE

SETBIT?CC macro Bitin,Val
if Bitin <32
  if Bitin > 23
k=3
  elseif  Bitin > 15
k=2
  elseif  Bitin > 7
k=1
  elseif  Bitin <8
k=0
  endif
b=Bitin-k*8


  if Val >0
  bsf _dis+k,b
  else
  bcf  _dis+k,b
  endif
endif
endm

not sure if user cmd is really useful here

[Ioannis]

Thanks Richard.

Interesting approach!

Ioannis

[Art]

I don’t yet know if I’ve wrecked it in basic.

shifter byte
offset byte
array byte[7]


shifter = 11
offset = shifter >> 3            // byte in array to pass to whatever needs the 4 byte result, in this case 2
array[3] = 1<<(shifter & 7)      // Set it to 00000000 (00000000 00000000 00001000 00000000) 00000000 00000000
‘dostuff with array[offset]

[Ioannis]

I could not follow that...

Ioannis

[Art]

So long as I have not ruined the way a C pointer will work...

Before any input, the 32 bits you want could be anywhere within the 7 byte array.
After computation, the memory location of the four bytes of the 32 bit value begins at array[offset].

[Ioannis]

Oh, I see. Thanks for the ideas.

Still, asm rotate will be faster.

Ioannis

[Art]

I have my doubts, You only have to call it once.
Assuming there is equal chance the input is 0-31, an assembler rotate routine will be called mean of 15 times.
For a shift in either language the overhead is in finding the byte you want to shift.

Again, assuming it’s still working. I’ll be able to try in PBP soon.

If I’m mistaken and it’s not another trick you’ve worked out because of the known state of the array at the beginning, that would be helpful.
It is currently taking me the equal number of assembler instructions as there are bytes in the array to bitwise rotate an array
that has unknown contents at the beginning, which is the same as in the sixth post of this thread.