Version analysis of LOIN (Part 2 of 7) / LOINQ (Part 2 of 7)

This code appears in the following versions (click to see it in the source code):

Cassette, Disc (flight), Disc (docked), 6502 Second Processor, Master and Electron

Code variations between these versions are shown below.

Code variation 1 of 10A variation in the comments only

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Cassette, Flight, Docked, 6502SP, Electron
Master
       Name: LOIN (Part 2 of 7)
       Name: LOINQ (Part 2 of 7)

       Type: Subroutine
   Category: Drawing lines
    Summary: Draw a line: Line has a shallow gradient, step right along x-axis
  Deep dive: Bresenham's line algorithm


 This routine draws a line from (X1, Y1) to (X2, Y2). It has multiple stages.
 If we get here, then:

   * |delta_y| < |delta_x|

   * The line is closer to being horizontal than vertical

   * We are going to step right along the x-axis

   * We potentially swap coordinates to make sure X1 < X2


.STPX

 LDX X1                 \ Set X = X1

 CPX X2                 \ If X1 < X2, jump down to LI3, as the coordinates are
 BCC LI3                \ already in the order that we want

 DEC SWAP               \ Otherwise decrement SWAP from 0 to &FF, to denote that
                        \ we are swapping the coordinates around

 LDA X2                 \ Swap the values of X1 and X2
 STA X1
 STX X2

 TAX                    \ Set X = X1

 LDA Y2                 \ Swap the values of Y1 and Y2
 LDY Y1
 STA Y1
 STY Y2

.LI3

                        \ By this point we know the line is horizontal-ish and
                        \ X1 < X2, so we're going from left to right as we go
                        \ from X1 to X2

Code variation 2 of 10Related to the screen mode

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Cassette, Flight, Docked
6502SP, Master
Electron
 LDA Y1                 \ Set A = Y1 / 8, so A now contains the character row
 LSR A                  \ that will contain our horizontal line
 LSR A
 LSR A

 ORA #&60               \ As A < 32, this effectively adds &60 to A, which gives
                        \ us the screen address of the character row (as each
                        \ character row takes up 256 bytes, and the first
                        \ character row is at screen address &6000, or page &60)

 STA SCH                \ Store the page number of the character row in SCH, so
                        \ the high byte of SC is set correctly for drawing the
                        \ start of our line
 LDY Y1                 \ Look up the page number of the character row that
 LDA ylookup,Y          \ contains the pixel with the y-coordinate in Y1, and
 STA SC+1               \ store it in SC+1, so the high byte of SC is set
                        \ correctly for drawing our line
                        \ We now calculate the address of the character block
                        \ containing the pixel (X1, Y1) and put it in SC(1 0),
                        \ as follows:
                        \
                        \   SC = &5800 + (Y1 div 8 * 256) + (Y1 div 8 * 64) + 32
                        \
                        \ See the deep dive on "Drawing pixels in the Electron
                        \ version" for details

 LDA Y1                 \ Set A = Y1 / 8, so A now contains the character row
 LSR A                  \ that will contain our horizontal line
 LSR A
 LSR A

 STA SC+1               \ Set SC+1 = A, so (SC+1 0) = A * 256
                        \                           = char row * 256

 LSR A                  \ Set (A SC) = (A SC) / 4
 ROR SC                 \            = (4 * ((char row * 64) + 32)) / 4
 LSR A                  \            = char row * 64 + 32
 ROR SC

 ADC SC+1               \ Set SC(1 0) = (A SC) + (SC+1 0) + &5800
 ADC #&58               \             = (char row * 64 + 32)
 STA SC+1               \               + char row * 256
                        \               + &5800
                        \
                        \ which is what we want, so SC(1 0) contains the address
                        \ of the first visible pixel on the character row
                        \ containing the point (X1, Y1)

 TXA                    \ Each character block contains 8 pixel rows, so to get
 AND #%11111000         \ the address of the first byte in the character block
                        \ that we need to draw into, as an offset from the start
                        \ of the row, we clear bits 0-2

 ADC SC                 \ And add the result to SC(1 0) to get the character
 STA SC                 \ block on the row we want

 BCC P%+4               \ If the addition of the low bytes overflowed, increment
 INC SC+1               \ the high byte

                        \ So SC(1 0) now contains the address of the first pixel
                        \ in the character block containing the (X1, Y1), taking
                        \ the screen borders into consideration

 LDA Y1                 \ Set Y = Y1 mod 8, which is the pixel row within the
 AND #7                 \ character block at which we want to draw the start of
 TAY                    \ our line (as each character block has 8 rows)

Code variation 3 of 10Related to the screen mode

This variation is blank in the Electron version.

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Cassette, Flight, Docked
6502SP, Master
 TXA                    \ Set A = bits 3-7 of X1
 AND #%11111000
 TXA                    \ Set A = 2 * bits 2-6 of X1
 AND #%11111100         \
 ASL A                  \ and shift bit 7 of X1 into the C flag

Code variation 4 of 10Related to the screen mode

This variation is blank in the Electron version.

Cassette, Flight, Docked, 6502SP, Master
 STA SC                 \ Store this value in SC, so SC(1 0) now contains the
                        \ screen address of the far left end (x-coordinate = 0)
                        \ of the horizontal pixel row that we want to draw the
                        \ start of our line on

Code variation 5 of 10Related to the screen mode

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Cassette, Flight, Docked, Electron
6502SP, Master
 TXA                    \ Set X = X1 mod 8, which is the horizontal pixel number
 AND #7                 \ within the character block where the line starts (as
 TAX                    \ each pixel line in the character block is 8 pixels
                        \ wide)

 LDA TWOS,X             \ Fetch a 1-pixel byte from TWOS where pixel X is set,
 STA R                  \ and store it in R
 BCC P%+4               \ If bit 7 of X1 was set, so X1 > 127, increment the
 INC SC+1               \ high byte of SC(1 0) to point to the second page on
                        \ this screen row, as this page contains the right half
                        \ of the row

 TXA                    \ Set R = X1 mod 4, which is the horizontal pixel number
 AND #3                 \ within the character block where the line starts (as
 STA R                  \ each pixel line in the character block is 4 pixels
                        \ wide)

Code variation 6 of 10Other (e.g. bug fix, optimisation)

Part 2 of the LOIN routine in the advanced versions uses logarithms to speed up the multiplication.

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Cassette, Flight, Docked, Electron
6502SP, Master
                        \ The following calculates:
                        \
                        \   Q = Q / P
                        \     = |delta_y| / |delta_x|
                        \
                        \ using the same shift-and-subtract algorithm that's
                        \ documented in TIS2

 LDA Q                  \ Set A = |delta_y|

 LDX #%11111110         \ Set Q to have bits 1-7 set, so we can rotate through 7
 STX Q                  \ loop iterations, getting a 1 each time, and then
                        \ getting a 0 on the 8th iteration... and we can also
                        \ use Q to catch our result bits into bit 0 each time

.LIL1

 ASL A                  \ Shift A to the left

 BCS LI4                \ If bit 7 of A was set, then jump straight to the
                        \ subtraction

 CMP P                  \ If A < P, skip the following subtraction
 BCC LI5

.LI4

 SBC P                  \ A >= P, so set A = A - P

 SEC                    \ Set the C flag to rotate into the result in Q

.LI5

 ROL Q                  \ Rotate the counter in Q to the left, and catch the
                        \ result bit into bit 0 (which will be a 0 if we didn't
                        \ do the subtraction, or 1 if we did)

 BCS LIL1               \ If we still have set bits in Q, loop back to TIL2 to
                        \ do the next iteration of 7

                        \ We now have:
                        \
                        \   Q = A / P
                        \     = |delta_y| / |delta_x|
                        \
                        \ and the C flag is clear

 LDX P                  \ Set X = P + 1
 INX                    \       = |delta_x| + 1
                        \
                        \ We add 1 so we can skip the first pixel plot if the
                        \ line is being drawn with swapped coordinates

 LDA Y2                 \ Set A = Y2 - Y1 - 1 (as the C flag is clear following
 SBC Y1                 \ the above division)

 BCS DOWN               \ If Y2 >= Y1 - 1 then jump to DOWN, as we need to draw
                        \ the line to the right and down
                        \ The following section calculates:
                        \
                        \   Q = Q / P
                        \     = |delta_y| / |delta_x|
                        \
                        \ using the log tables at logL and log to calculate:
                        \
                        \   A = log(Q) - log(P)
                        \     = log(|delta_y|) - log(|delta_x|)
                        \
                        \ by first subtracting the low bytes of the logarithms
                        \ from the table at LogL, and then subtracting the high
                        \ bytes from the table at log, before applying the
                        \ antilog to get the result of the division and putting
                        \ it in Q

 LDX Q                  \ Set X = |delta_y|

 BEQ LIlog7             \ If |delta_y| = 0, jump to LIlog7 to return 0 as the
                        \ result of the division

 LDA logL,X             \ Set A = log(Q) - log(P)
 LDX P                  \       = log(|delta_y|) - log(|delta_x|)
 SEC                    \
 SBC logL,X             \ by first subtracting the low bytes of log(Q) - log(P)

Code variation 7 of 10Other (e.g. bug fix, optimisation)

The Master version omits half of the logarithm algorithm when compared to the 6502SP version.

See below for more variations related to this code.

This variation is blank in the Cassette, Disc (flight), Disc (docked), Master and Electron versions.

6502SP
 BMI LIlog4             \ If A > 127, jump to LIlog4

Code variation 8 of 10Other (e.g. bug fix, optimisation)

See variation 7 above for details.

This variation is blank in the Cassette, Disc (flight), Disc (docked) and Electron versions.

Tap on a block to expand it, and tap it again to revert.

Master
6502SP
 LDX Q                  \ And then subtracting the high bytes of log(Q) - log(P)
 LDA log,X              \ so now A contains the high byte of log(Q) - log(P)
 LDX P
 SBC log,X

 BCS LIlog5             \ If the subtraction fitted into one byte and didn't
                        \ underflow, then log(Q) - log(P) < 256, so we jump to
                        \ LIlog5 to return a result of 255

 TAX                    \ Otherwise we set A to the A-th entry from the antilog
 LDA alogh,X            \ table so the result of the division is now in A

 JMP LIlog6             \ Jump to LIlog6 to return the result

.LIlog5

 LDA #255               \ The division is very close to 1, so set A to the
 BNE LIlog6             \ closest possible answer to 256, i.e. 255, and jump to
                        \ LIlog6 to return the result (this BNE is effectively a
                        \ JMP as A is never zero)

.LIlog7
 LDX Q                  \ And then subtracting the high bytes of log(Q) - log(P)
 LDA log,X              \ so now A contains the high byte of log(Q) - log(P)
 LDX P
 SBC log,X

 BCS LIlog5             \ If the subtraction fitted into one byte and didn't
                        \ underflow, then log(Q) - log(P) < 256, so we jump to
                        \ LIlog5 to return a result of 255

 TAX                    \ Otherwise we set A to the A-th entry from the antilog
 LDA antilog,X          \ table so the result of the division is now in A

 JMP LIlog6             \ Jump to LIlog6 to return the result

.LIlog5

 LDA #255               \ The division is very close to 1, so set A to the
 BNE LIlog6             \ closest possible answer to 256, i.e. 255, and jump to
                        \ LIlog6 to return the result (this BNE is effectively a
                        \ JMP as A is never zero)

.LIlog7

Code variation 9 of 10Other (e.g. bug fix, optimisation)

See variation 7 above for details.

This variation is blank in the Cassette, Disc (flight), Disc (docked) and Electron versions.

Tap on a block to expand it, and tap it again to revert.

Master
6502SP
 LDA #0                 \ The numerator in the division is 0, so set A to 0
 LDA #0                 \ The numerator in the division is 0, so set A to 0 and
 BEQ LIlog6             \ jump to LIlog6 to return the result (this BEQ is
                        \ effectively a JMP as A is always zero)

.LIlog4

 LDX Q                  \ Subtract the high bytes of log(Q) - log(P) so now A
 LDA log,X              \ contains the high byte of log(Q) - log(P)
 LDX P
 SBC log,X

 BCS LIlog5             \ If the subtraction fitted into one byte and didn't
                        \ underflow, then log(Q) - log(P) < 256, so we jump to
                        \ LIlog5 to return a result of 255

 TAX                    \ Otherwise we set A to the A-th entry from the
 LDA antilogODD,X       \ antilogODD so the result of the division is now in A

Code variation 10 of 10Other (e.g. bug fix, optimisation)

See variation 7 above for details.

This variation is blank in the Cassette, Disc (flight), Disc (docked) and Electron versions.

6502SP, Master
.LIlog6

 STA Q                  \ Store the result of the division in Q, so we have:
                        \
                        \   Q = |delta_y| / |delta_x|

 LDX P                  \ Set X = P
                        \       = |delta_x|

 BEQ LIEXS              \ If |delta_x| = 0, return from the subroutine, as LIEXS
                        \ contains a BEQ LIEX instruction, and LIEX contains an
                        \ RTS

 INX                    \ Set X = P + 1
                        \       = |delta_x| + 1
                        \
                        \ We add 1 so we can skip the first pixel plot if the
                        \ line is being drawn with swapped coordinates

 LDA Y2                 \ If Y2 < Y1 then skip the following instruction
 CMP Y1
 BCC P%+5

 JMP DOWN               \ Y2 >= Y1, so jump to DOWN, as we need to draw the line
                        \ to the right and down