GARTHWILSON wrote:
FE + 02 = 00 if C had been clear or 01 if C had been set, and C ends up set either way;
FE + 04 = 02 if C had been clear or 03 if C had been set, and C ends up set either way;
FE + 08 = 06 if C had been clear or 07 if C had been set, and C ends up set either way;
etc., and:
FE + 00 = FE if C had been clear or FF if C had been set, and C ends up clear either way.
Thank goodness! I thought that's what you meant, but you've explained it much better than I could. Still, I don't quite see how it'll gain us anything in this case.
Quote:
[...] (if you're going to waste the rest of the port anyway). Another possibility is to make the other bits outputs [...]
The routine is intended to allow you to use any bit -- even bit 0 -- for the serial input, disregarding any activity (including outputs) on the other bits. I
think I succeeded in that... Here it is with the comments (hopefully) improved!
Code:
ComBit equ $01 ;To determine the serial data input, change ComBit as required and reassemble.
;In this example the input is on bit 0.
?RcvByte lda #ComBit ; 2~ A= 01 ie; the ComBit
?r bit portb ; 4~ read portb and test ComBit
beq ?r ; 2~ fall thru only when the Start Bit has been detected
lda #$40 ; 2 Seed the data shift register so it can
sta pom ; 4 count the incoming bits.(No need to use X or Y for counting.)
#cycle #8 ; 8
?Loop ; the loop inputs the first 7 bits
lda #NOT(ComBit) ; 2 A= fe
ora portb ; 4 A= fe else A= ff
cmp #$ff ; 2 fe-ff clears C else ff-ff sets C
ror pom ; 6 Use Absolute addressing mode! (for timing)
bcc ?Loop ; 3/2
#cycle #2 ; 2
lda #NOT(ComBit) ; 2 A= fe (etc as above)
ora portb ; 4
cmp #$ff ; 2
lda pom ; 4 Retrieve first 7 bits
ror @ ; 2 ror Accumulator
eor #$ff ; 2 Invert the bits
rts
Quote:
If you use bit 0, then you'd have to CLC before adding
Right; 65xx CPUs don't offer an "Add
Without Carry" instruction. In order to omit the CLC and still get exact arithmetic I switched to using CMP instead -- which is "
Subtract Without Carry" (aka /Borrow).
Quote:
If the byte is rotated into a memory location instead of A, it takes more cycles
True, of course, and the final rotate does occur in A. The interesting thing about this sort of routine is that a certain amount of delay (between input samples) is mandatory and cannot be "optimized." However, what you
can do is reorder operations as much as possible so there's a minimum of processing
before the first input sample and
after the last. As
leeeeee noted, it's beneficial to allow as many extra cycles as possible to do something with the byte once you have it.
-- Jeff
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(#cycle #2 after escaping the loop may seem excessive, but #cycle #1 is not an option. Things will be ahead of schedule here anyway, because we ideally should allow 17.36~ per bit, not 17.)
Cheers,
, so I appreciate the pedantic devil in you