Another way to left shift a 32 bit data?

[HenrikOlsson]

How about the DCD operator? Is that what you're looking for? Is it fast enough?

DCD returns the decoded value of a bit number. It changes a bit number (0 - 31) into a binary number with only that bit set to 1. All other bits are set to 0.

B0 = DCD 2  ' Sets B0 to %00000100

/Henrik.

[Ioannis]

Thanks for the ideas!

About the time they take, I have to test it. But the George's idea I guess it needs a fixed time. The DCD may need different time for setting the first or last bit using a loop.

Thanks for the replies.

Ioannis

[Art]

Where did you get a 32 bit variable that PBP knows about? Is this a feature of PBP3?
Can you alias and access the same variable as the four bytes?
Since towerg mentioned “array” I’m not sure. a single variable of a datatype isn’t an array.

[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