.STPY LDY Y1 \ Set A = Y = Y1 TYA LDX X1 \ Set X = X1 CPY Y2 \ If Y1 >= Y2, jump down to LI15, as the coordinates are BCS LI15 \ 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 STA Y1 STY Y2 TAY \ Set Y = A = Y1 .LI15 \ By this point we know the line is vertical-ish and \ Y1 >= Y2, so we're going from top to bottom as we go \ from Y1 to Y2 LSR A \ Set A = Y1 / 8, so A now contains the character row LSR A \ that will contain our horizontal line 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 TXA \ Set A = bits 3-7 of X1 AND #%11111000 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 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 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) \ The following calculates: \ \ P = P / Q \ = |delta_x| / |delta_y| \ \ using the same shift-and-subtract algorithm \ documented in TIS2 LDA P \ Set A = |delta_x| LDX #1 \ Set Q to have bits 1-7 clear, so we can rotate through STX P \ 7 loop iterations, getting a 1 each time, and then \ getting a 1 on the 8th iteration... and we can also \ use P to catch our result bits into bit 0 each time .LIL4 ASL A \ Shift A to the left BCS LI13 \ If bit 7 of A was set, then jump straight to the \ subtraction CMP Q \ If A < Q, skip the following subtraction BCC LI14 .LI13 SBC Q \ A >= Q, so set A = A - Q SEC \ Set the C flag to rotate into the result in Q .LI14 ROL P \ Rotate the counter in P 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) BCC LIL4 \ If we still have set bits in P, loop back to TIL2 to \ do the next iteration of 7 \ We now have: \ \ P = A / Q \ = |delta_x| / |delta_y| \ \ and the C flag is set LDX Q \ Set X = Q + 1 INX \ = |delta_y| + 1 \ \ We add 1 so we can skip the first pixel plot if the \ line is being drawn with swapped coordinates LDA X2 \ Set A = X2 - X1 (the C flag is set as we didn't take SBC X1 \ the above BCC) BCC LFT \ If X2 < X1 then jump to LFT, as we need to draw the \ line to the left and downName: LOIN (Part 5 of 7) [Show more] Type: Subroutine Category: Drawing lines Summary: Draw a line: Line has a steep gradient, step up along y-axis Deep dive: Bresenham's line algorithmContext: See this subroutine in context in the source code Variations: See code variations for this subroutine in the different versions References: No direct references to this subroutine in this source file
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 vertical than horizontal * We are going to step up along the y-axis * We potentially swap coordinates to make sure Y1 >= Y2
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Label LFT in subroutine LOIN (Part 7 of 7)
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Label LI13 is local to this routine
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Label LI14 is local to this routine
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Label LI15 is local to this routine
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Label LIL4 is local to this routine
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Variable TWOS (category: Drawing pixels)
Ready-made single-pixel character row bytes for mode 4