I wanna see how fast and small you can make some code to turn a 16 bit register backwards
%1101110 01101010
becomes
%01010110 0111011
I did it in 40 bytes and 156 clocks.
I'll post the code tomorrow.
oh, and whoever wins posts the next challenge.
Show us. I want to see how you got it to 100 clocks
It's accurate. Your IDE probably has a code counter feature.
What IDE? I'm just counting code. I'm double checking it now.
use wabbitcode for code counting. or use asm8x or something.
wabbitcode is easiest, just highlight it and look at the bottom right corner of the window.
Wait, its 154 clocks. My bad. And the size is 38 bytes. Still win.
hl reversed to hl in 30 bytes and 120 clocks:
Code: ld a,l ;one byte and 4 clocks
rr h
rla ;bits 7 and 8; three bytes and 4+8=12 clocks
rr h
rla ;bits 6 and 9; three bytes and 12 clocks
rr h
rla ;bits 5 and 10; three bytes and 12 clocks
rr h
rla ;bits 4 and 11; three bytes and 12 clocks
rr h
rla ;bits 3 and 12; three bytes and 12 clocks
rr h
rla ;bits 2 and 13; three bytes and 12 clocks
rr h
rla ;bits 1 and 14; three bytes and 12 clocks
rr h
rla ;bits 0 and 15; three bytes and 12 clocks
rr h ;catch the carry out of the final bit of a=l; 2b, 8c
rlc h ;rotate it properly
ld l,a ;reassemble; 1b, 4c
KermMartian wrote:
hl reversed to hl in 30 bytes and 120 clocks:
Code: ld a,l ;one byte and 4 clocks
rr h
rla ;bits 7 and 8; three bytes and 4+8=12 clocks
rr h
rla ;bits 6 and 9; three bytes and 12 clocks
rr h
rla ;bits 5 and 10; three bytes and 12 clocks
rr h
rla ;bits 4 and 11; three bytes and 12 clocks
rr h
rla ;bits 3 and 12; three bytes and 12 clocks
rr h
rla ;bits 2 and 13; three bytes and 12 clocks
rr h
rla ;bits 1 and 14; three bytes and 12 clocks
rr h
rla ;bits 0 and 15; three bytes and 12 clocks
rr h ;catch the carry out of the final bit of a=l; 2b, 8c
rlc h ;rotate it properly
ld l,a ;reassemble; 1b, 4c
38 bytes/154 cycles
Code: ; Reverse bits of 2 bytes
; Input: hl points to first byte
; a and b are destroyed.
; Output: hl points second byte of set.
Reverse16Bit
ld a, (hl)
call reverseA
ld b, a
inc hl
ld a, (hl)
call reverseA
dec hl
ld (hl), a
ld a, b
inc hl
ld (hl), a
ret
reverseA:
ld b,a
rrca
rrca
xor b
and %10101010
xor b
ld b,a
rrca
rrca
rrca
rrca
xor b
and %01100110
xor b
rrca
call Reverse16Bit
Code: di
ld a,h
ex af, af'
ld a,l
rrca
rrca
xor l
and %10101010
xor l
ld l,a
rrca
rrca
rrca
rrca
xor l
and %01100110
xor l
rrca
ex af, af'
rrca
rrca
xor h
and %10101010
xor h
ld h,a
rrca
rrca
rrca
rrca
xor h
and %01100110
xor h
rrca
ld l,a
ex af, af'
ld h,a
Second place isn't horrible
SirCmpwn, you forgot that reverseA is executed twice. And you miscounted.
Edit: and I got pwnt by the most obvious solution ever. Guess I was trying to be too fancy.
Nicely done, guys.
Here's a smaller and slower version of mine (11 bytes and 196 clocks):
Code: ld a,l ;one byte and 4 clocks
ld b,8 ;two bytes and 7 clocks
repeat:
rr h
rla ;three bytes and 4+8=12 clocks
djnz repeat ;8*7+13 clocks, 2 bytes
rlc h ;rotate it properly
ld l,a ;reassemble; 1b, 4c
Grr, whats the real one? Its quite a bit harder to count by hand.
And KermM - holy crap
28 bytes, 112 cycles
Code: ld a,l
rla
rr h
rla
rr h
rla
rr h
rla
rr h
rla
rr h
rla
rr h
rla
rr h
rla
rr h
rla
rrca
ld l,a
I win 
Fast and small are most often mutually exclusive.
Pick the more important one next time.
calc84maniac wrote:
28 bytes, 112 cycles
Code: ld a,l
rla
rr h
rla
rr h
rla
rr h
rla
rr h
rla
rr h
rla
rr h
rla
rr h
rla
rr h
rla
rrca
ld l,a
I win
Pwnt
Edit: also its your Turn Calc84 to make a challenge.
Hey, all you did was use my code slightly permuted.
Nice job, calc84.
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