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a question for the ASM black belts in here:
how would you swap position of two bits in a byte quickly without using a LUT?

I mean, say I want %abcdefgh to become %abgdefch

if it's easier, I may settle for a solution that works only if the two bits are adjacent, as in %abcedfgh

replying myself - I probably have an idea that involves testing the two bits and a XOR operation to be possibly performed

will leave this open for now ;)

Untested...

; test value is in b, bitmask for the bits in question in c
; return in a
ld a,b
and c
jr z, +
cp c
jr z, +
ld a,b
xor c ; we are flipping the bits
ret
+:
; the bits are identical
ld a,b
ret

The idea is that if the bits are the same, there's nothing to do, else you want to XOR them with the bitmask. If they can't be the same, or the bitmask is constant, there may be some possibility to optimise.

yeah, I had the very same idea, but then I thought we can actually do the two separate test as one

; test value is in b, bitmask for the bits in question in c
; return in a
ld a,b
and c
ld a,b ; this doesn't affect flags
jp pe,+ ; check if Parity is EVEN
xor c ; if not, we are flipping the bits
+:
ret

Ah, that’s a good one... the parity flag is so rarely useful that I forgot it :)

I think you can ret pe to save a bit more time and space.

improved version. Thanks Maxim - I had forgot about the conditional RET

; test value is in b, bitmask for the bits in question in c
; return in a
ld a,b
and c
ld a,b ; this doesn't affect flags
ret pe ; if Parity Flag is EVEN we don't need any flip
xor c ; if not, we are flipping the bits
ret

Author	Message
sverx Joined: 05 Sep 2013 Posts: 3811 Location: Stockholm, Sweden	swap two bits in a byte Posted: Fri Mar 13, 2020 11:13 am
	a question for the ASM black belts in here: how would you swap position of two bits in a byte quickly without using a LUT? I mean, say I want %abcdefgh to become %abgdefch if it's easier, I may settle for a solution that works only if the two bits are adjacent, as in %abcedfgh

sverx Joined: 05 Sep 2013 Posts: 3811 Location: Stockholm, Sweden	Posted: Fri Mar 13, 2020 11:24 am
	replying myself - I probably have an idea that involves testing the two bits and a XOR operation to be possibly performed will leave this open for now ;)

Maxim Site Admin Joined: 19 Oct 1999 Posts: 14732 Location: London	Posted: Fri Mar 13, 2020 11:34 am
	Untested... ; test value is in b, bitmask for the bits in question in c ; return in a ld a,b and c jr z, + cp c jr z, + ld a,b xor c ; we are flipping the bits ret +: ; the bits are identical ld a,b ret The idea is that if the bits are the same, there's nothing to do, else you want to XOR them with the bitmask. If they can't be the same, or the bitmask is constant, there may be some possibility to optimise.

sverx Joined: 05 Sep 2013 Posts: 3811 Location: Stockholm, Sweden	Posted: Fri Mar 13, 2020 11:44 am
	yeah, I had the very same idea, but then I thought we can actually do the two separate test as one ; test value is in b, bitmask for the bits in question in c ; return in a ld a,b and c ld a,b ; this doesn't affect flags jp pe,+ ; check if Parity is EVEN xor c ; if not, we are flipping the bits +: ret

Maxim Site Admin Joined: 19 Oct 1999 Posts: 14732 Location: London	Posted: Fri Mar 13, 2020 11:46 am
	Ah, that’s a good one... the parity flag is so rarely useful that I forgot it :) I think you can ret pe to save a bit more time and space.

sverx Joined: 05 Sep 2013 Posts: 3811 Location: Stockholm, Sweden	Posted: Fri Mar 13, 2020 11:54 am
	improved version. Thanks Maxim - I had forgot about the conditional RET ; test value is in b, bitmask for the bits in question in c ; return in a ld a,b and c ld a,b ; this doesn't affect flags ret pe ; if Parity Flag is EVEN we don't need any flip xor c ; if not, we are flipping the bits ret

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