Shift bits left in .asm for PIC18

Question

I'm trying to shift bits left in .asm via the rlcf by rotating, and then clearing the carry. Is this going to work? If not, can I get some help? For some reason this won't compile and it's driving me crazy, this is such a simple program yet I can't get it. All help greatly appreciated.

count equ 0x20
byte equ 0x21

;shift right

movlw b'11111110'
movwf byte
bcf STATUS,C
rlcf Byte ;for left shift
bcf STATUS,C
loop
end

score 1 · Answer 1 · answered Oct 16 '12 at 20:29

1

Note that shift left X by 1 is exactly X + X -- use the add (addwf) instruction.

answered Oct 16 '12 at 20:29

Doug Currie

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we're required to use the rotate function, which essentially should be the same, correct? – matt Oct 16 '12 at 20:32
hm, should work then maybe im not loading settings or something. thank you – matt Oct 16 '12 at 20:34
1

See Paul R's comment as well -- the assembler is case sensitive. – Doug Currie Oct 16 '12 at 20:35

Shift bits left in .asm for PIC18

1 Answers1