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Rudi
Account closed
Registered: May 2010
Posts: 125 |
Fast way to rotate a char?
Im not talking about rol or ror, but swap bits so that they are rotated 90 degrees:
Example:
a char (and the bits can be random):
10110010 byte 1..
11010110 byte 2.. etc..
00111001
01010110
11011010
10110101
00110011
10110100 after "rotation" (rows and columns are swapped):
11001101
01011000
10100111
11111111
00101000
01010101
11011010
00100110 is it possible to use lookup tables for this or would that lookup table be too big?
or other lookuptable for getting and setting bits?
-Rudi |
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Axis/Oxyron
Registered: Apr 2007
Posts: 91 |
Rudi, except of the fact that you used EOR instead of ORA and didnt use LAX for the first read on zp (would be possible if you swap x and y registers), this looks identical to the code I posted pretty early in this thread. Is there a special reason to use EOR? |
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Rudi
Account closed
Registered: May 2010
Posts: 125 |
Quote: Rudi, except of the fact that you used EOR instead of ORA and didnt use LAX for the first read on zp (would be possible if you swap x and y registers), this looks identical to the code I posted pretty early in this thread. Is there a special reason to use EOR?
The eor was a consequence of the formulas i used for masking and swapping (I derived this from Kalms tutor):
Example for the 4x4 swapping:tmp0 = byte0 & 0xf0; //xxxx----
tmp1 = byte1 & 0xf0; //xxxx----
tmp2 = byte2 & 0xf0; //xxxx----
tmp3 = byte3 & 0xf0; //xxxx----
tmp4 = byte4 & 0x0f; //----xxxx
tmp5 = byte5 & 0x0f; //----xxxx
tmp6 = byte6 & 0x0f; //----xxxx
tmp7 = byte7 & 0x0f; //----xxxx
data0 = byte0 << 4;
data1 = byte1 << 4;
data2 = byte2 << 4;
data3 = byte3 << 4;
data4 = byte4 >> 4;
data5 = byte5 >> 4;
data6 = byte6 >> 4;
data7 = byte7 >> 4;
data0 ^= tmp4;
data1 ^= tmp5;
data2 ^= tmp6;
data3 ^= tmp7;
data4 ^= tmp0;
data5 ^= tmp1;
data6 ^= tmp2;
data7 ^= tmp3; Sorry for the long code..
EOR is used for the last xor-swapping. Since I cannot use lookup-table for two different values (fex. data0 and tmp4). The last eight operations in the above are done with the EOR-instruction.
Some of the lookup-tables i derived are doing EOR, AND and SHIFTS at the same time:shl2_eor_cc[i] = (i ^ (i & 0xcc)) << 2;
shr2_eor_33[i] = (i ^ (i & 0x33)) >> 2;
shl1_eor_aa[i] = (i ^ (i & 0xaa)) << 1;
shr1_eor_55[i] = (i ^ (i & 0x55)) >> 1; I scratched my head around how your ORA worked. And since I didnt understand that Dreamass-macrocode I wrote mine from scratch. But maybe I should look at your LAX-method next.
Edit: Now I see that it doesnt really matter if one use ora or eor for this technique. |
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Axis/Oxyron
Registered: Apr 2007
Posts: 91 |
But you know that:
(i ^ (i & 0xcc))
is the same as:
i & 0x33
;o) |
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Rudi
Account closed
Registered: May 2010
Posts: 125 |
Quote: But you know that:
(i ^ (i & 0xcc))
is the same as:
i & 0x33
;o)
No, didnt think about that hehe.
Btw, 312 cycles now (with LAX). |
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Bitbreaker
Registered: Oct 2002
Posts: 500 |
Quoting Axis/OxyronBut you know that:
(i ^ (i & 0xcc))
is the same as:
i & 0x33
;o)
smells like the version of the tab that shifts only 1 bit to the right could be substituted by some asr magic?
Also the and maskX looks like it could be included into something, too static to be done that often :-) |
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Rudi
Account closed
Registered: May 2010
Posts: 125 |
XAA might be something too. |
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Rudi
Account closed
Registered: May 2010
Posts: 125 |
Here's a different approach to it:ldx $82 ;3
xaa #$33 ;2 a=(x & 0x33)
ldy $80 ;3
eor shl2_eor_cc, y ;4*
sta $90 ;3
lda shr2_eor_33, x ;4*
eor tab_cc, y ;4*
sta $92 ;3 uses the same amount of cycles though.
Bitbreaker: yes, one could probably optimize the 1x1 rotator with other illegal-opcodes. sine some of them do one shift. |
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Bitbreaker
Registered: Oct 2002
Posts: 500 |
Besides that it will produce rubbish as xaa can add some unpredictable value to A before doing the txa and and part :-) |
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Rastah Bar
Registered: Oct 2012
Posts: 336 |
Quoting Color BarI may have found a method that takes 432 cycles....
If I merge columns of 2 bits wide and 4 bits high into one byte and then extract the destination nybbles I can reduce that to 354 cycles. |
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Rudi
Account closed
Registered: May 2010
Posts: 125 |
Quote: Quoting Color BarI may have found a method that takes 432 cycles....
If I merge columns of 2 bits wide and 4 bits high into one byte and then extract the destination nybbles I can reduce that to 354 cycles.
Are you using the masking method? |
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