Another way to left shift a 32 bit data?

[Art]

You didn’t say what you want to do with the bit that drops off the end.
Assuming you want what PBP shift would do with a byte, and just drop the end bit,
and feed in a zero to the right.

dis var byte [4]

BitwiseRotateLeft:						' bitwise rotate array
  @ rlf 	_dis+3		,F				; ditching the first bit
  @ rlf		_dis+2		,F				;
  @ rlf		_dis+1		,F				;
  @ rlf		_dis+0		,F				;
  @ bcf		_dis+3		,0				;clear MSB
return

So long as you can access the first byte by aliasing it, I’m sure the assembler is a continuous 4 bytes.

[Ioannis]

Well, yes, it is a PBP3 feature. But besides that, even if you are on a previous version, you may still face the fact of 32 bit variables.

At the cost of 4x8x32 bytes, George's approach is obviously the fastest because n (times a shiftmust happen) can be anywhere between 1 and 32.

Thanks for the post.

Ioannis

[Art]

Oh I get it, the 32 bit var always has 31 clear bits and 1 set bit.

[Ioannis]

Yes, thats right.

In case you wonder is part of the Microchip Keeloq algo. A really intense algo...

Ioannis

[Art]

Can LOOKUP/2 now contain enough elements,
or you'd need to use 4 lookups? (One for each byte element of the 32 bit val).
It does sound like reading program memory would be economical for that.

[Ioannis]

Since using LONGs in the program will slow down the execution of the program, I am reluctant to use longs. Also the PIC that curretly is used is 16F series.

So, seems better to spend some bytes in RAM or FLASH and have immediate response.

Ioannis

[Art]

That’s what I mean. Still a more difficult problem than appears, or more than it intuitively feels like it should be, if speed is the goal.
The extra 6 bits returned by a single 14 bit READCODE are useless to you because of the time taken to extract them.
The word variable you get to use to address program memory is also useless to you because you only needed a 32 bit index (or 32x4).
I’d be interested to hear how you end up tackling it.