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MaxVT103
Member
Joined: 24 Aug 2003
Posts: 109
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 Posted: 05 Nov 2003 05:17:55 pm Post subject: |
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Ok, does anyone know of any good tutorials or know how to read from a file for a regular asm program.
Not so wise Duck  |
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Jeffrey
Member
Joined: 12 Jun 2003
Posts: 212
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| Posted: 05 Nov 2003 07:47:46 pm Post subject: |
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I didnt write this, Justin Wales did (by the way I figured out what my problem was with reading the numbers, Justin)
Code:
;Standard template for ti83+ asm programs
#include "ti83plus.inc";File needed to access system routines
.org userMem-2 ;Define where to start in memory
.db $BB,$6D ;AsmPrgm instruction
ld hl,appvarname
rst 20h ;don't use bcall(_Mov9ToOp1) it takes 2 more bytes than rst 20h and they
;do the same thing.
ld hl,24
bcall(_CreateAppVar)
;The ROM call CreateAppVar once executed will return the same pointers as ChkFindSym therefore we
;do not need to use ChkFindSym in this case as we can immediately start writing to the appvar
inc de
inc de;by incrementing de twice we have bypassed the size bytes of the appvar
push de;save this pointer for later
ld hl,stufftoput ;load pointer to text into hl
ld bc,10 ;10 being the length of the string
ldir ;copy the bytes
bcall(_getkey)
bcall(_clrlcdfull)
;ok here's where I teach you a bit of optimization
;as you remember I pushed de above to save a pointer
;well that pointer is then stored to the stack
;the neat thing is, we don't have to pop that number back into de.
;we can pop it back into any 16 bit register we want, therefore if we pop it back into
;hl we no longer need the instruction ex de,hl to switch the pointers
;this results in a 1 byte smaller program and takes 4 clock cycles less per execution.
;these optimizations are rather small and insignificant in this small program but it's things
;like this that add up in the long run.
pop hl
;there is another major optimization in the next few lines.
;first of all, there is no need for 2 xor a instructions as a doesn't change
;in the ld (curcol),a instruction
;
;second currow preceeds curcol in RAM. Therefore we can use a 16 bit load instruction to load
;both coordinates in two instructions instead of 4
ld de,00*256+00 ;the first 2 0's are for the cursor row and the 2nd 2 are for the column ld (currow),de ;load 1st coordinate to currow, the other byte is copied to curcol
;the method you used takes 8 bytes and 34 clock cycles
;mine takes 7 bytes and 30 clock cycles
;please note you could use this as well
;
; xor a
; ld (currow),a
; ld (curcol),a
;
;That takes the same amount of clock cycles and bytes as my routine above but remember
;that this would only work if both coordinates were 0. My method is more effective overall
bcall(_puts)
bcall(_getkey)
ret
stufftoput:
.db "Test Text",0
appvarname:
.db AppVarObj,"Tstr",0
.end
;I hope I haven't cluttered your file too much, just thought I'd help ya out a bit :)
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