So, the best option here would be, im assuming, to just make a separate sprite for each directional position?

Can someone write me up one if they have the time? I started to write one myself, but got confused. Below is what I have so far. Can someone please help me out with this. I'm looking for the routine to prompt for an equation, allow you to press the down arrow for subscripts, make the second character of a two-character symbol lower-case, and make the y= button give the arrow.

ReceiveEquation:
    bcall(_ClrLCDFull)
    ld  a,0
    ld  (penCol),a
    ld  (penRow),a
    ld  de,saferam1+1
    ld  a,0
    ld  (saferam1),a
    ld  a,0
getKloop:
    bcall(_getCSC)
    jr  z,getKloop
    cp  $01
    jr  z,switchnumeric
    cp  $04
    jr  z,switchalpha
    cp  $02
    jr  z,backspace
    cp  $09
    jr  z,done


switchnumeric:
    bit 0,(TextInputFlag)
    jr  z,getKloop
    ld  hl,NumericCharMap
    ld  a,(penRow)
    add a,3
    ld  (penRow),a
    res 0,(TextInputFlag)
    jr  getKloop

switchalphaup:
    bit 0,(TextInputFlag)
    jr  nz,getKloop
    ld  hl,AlphaCharMap
    ld  a,(penRow)
    sub 3
    ld  (penRow),a
    set 0,(TextInputFlag)
    jr  getKloop

backspace:
    ld  a,(saferam1)
    cp  0
    jr  z,getKloop
    dec a
    ld  (saferam1),a
    dec de
    ld  a,(penCol)
    sub 5
    ld  (penCol),a
    jr  getKloop

done:
       
AlphaUpCharMap:
    .db 0,0,0,0,0,0,0,0,0,0
    .db "WRMH"
AlphaLowCharMap:

NumericCharMap:


thepenguin, a great job. However, one issue of convenience.

I lose the game (or win). I then get the cursor, but am unable to move. The game should automatically quit and return you to the main menu. At current, you need to exit manually, then it asks you if you are sure you want to exit a game that is already over.

Code: [Select]

// Multiply a times b
temp = 0
repeat for each bit in a
temp <<= 1
if (high bit of a set)
temp += b
a <<= 1
return temp
if a and b are 2 bytes, temp is 4 bytes, and you loop 16 times.

Spoiler For for code:

stolen from Axe
p_MulFull:
   ; Input in hl, result in cahl
   ld   c,h
   ld   a,l
   ld   hl,0   ;11
   ld   b,16   ;7
__MulFullNext:
   add   hl,hl   ;11
   rla      ;4
   rl   c   ;8
   jr   nc,__MulFullSkip   ;12/7
   add   hl,de   ;11
   adc   a,0   ;7
   jr   nc,__MulFullSkip
   inc   c
__MulFullSkip:
   djnz   __MulFullNext
   ret
__MulFullEnd:

Code: [Select]

// Sqrt a
temp = high byte of a
a <<= 8
b = 0
repeat for every 2 bits in a
test = b << 8 + 0x40
b <<= 1
if (temp >= test)
temp -= test
set low bit of b
temp += high 2 bits of a
a <<= 2
return b
If a is 4 bytes, then b and temp are 2 bytes, and you loop 16 times.

Spoiler For code:

stole my own routine from axe (and modified it)
p_Sqrt88:
   ; input in hlde, result in de
   ld   b,16
   ld   a,h
   ld   c,l
   push   de ; ld ixh,d
   pop   ix ; ld ixl,e
   ld   de,0
   ld   h,d
   ld   l,e
__Sqrt88Loop:
   sub   $40
   sbc   hl,de
   jr   nc,__Sqrt88Skip
   add   a,$40
   adc   hl,de
__Sqrt88Skip:
   ccf
   rl   e
   rl   d
   add   ix,ix
   rl   c
   rla
   adc   hl,hl
   add   ix,ix
   rl   c
   rla
   adc    hl,hl
   djnz   __Sqrt88Loop
   ret
__Sqrt88End:

Kool. Thanks. But, shouldn't the first one have two inputs?

Ok, I think I get it.

You move through the second register. If the current bit is 1, you add 1 x a to a, then rla a. If the current bit is 0, just rla a.

Ok, I'm actually going to be sticking with just two bytes for position and scrapping the sector. That means I'll need to do 2-byte subtraction (no problem), square a 2-byte (rendering a max 4-byte), then add three 4-bytes (do you only need a 4-byte output or should you go up to 5), then square root that. That should be easier.

√( (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)

√( (16bit-16bit)^2 + (16bit-16bit)^2 + (16bit-16bit)^2)


Can someone explain to me the theory behind multiplying/dividing/sqrt'ing numbers? Is there a standard theorem, regardless of the bit-size of the number?

use thepenguin's patch. What is your OS #?

Ideas? Anyone willing to have "fun" with this?