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Elite on the BBC Micro and NES

Drawing circles: CIRCLE2

[6502 Second Processor version]

Name: CIRCLE2 [Show more] Type: Subroutine Category: Drawing circles Summary: Draw a circle (for the planet or chart) Deep dive: Drawing circles
Context: See this subroutine in context in the source code Variations: See code variations for this subroutine in the different versions References: This subroutine is called as follows: * HFS2 calls CIRCLE2 * TT128 calls CIRCLE2

Draw a circle with the centre at (K3, K4) and radius K. Used to draw the planet and the chart circles.
Arguments: STP The step size for the circle K The circle's radius K3(1 0) Pixel x-coordinate of the centre of the circle K4(1 0) Pixel y-coordinate of the centre of the circle
Returns: C flag The C flag is cleared
Other entry points: CIRCLE3 Just add the circle segments to the existing ball line heap - do not send the send the ball line heap to the I/O processor for drawing on-screen
.CIRCLE3 \ This gets called from CIRCLE2 below to calculate the \ line segments, which CIRCLE2 then sends to the I/O \ processor for drawing LDX #&FF \ Set FLAG = &FF to reset the ball line heap in the call STX FLAG \ to the BLINE routine below INX \ Set CNT = 0, our counter that goes up to 64, counting STX CNT \ segments in our circle .PLL3 LDA CNT \ Set A = CNT JSR FMLTU2 \ Call FMLTU2 to calculate: \ \ A = K * sin(A) \ = K * sin(CNT) LDX #0 \ Set T = 0, so we have the following: STX T \ \ (T A) = K * sin(CNT) \ \ which is the x-coordinate of the circle for this count LDX CNT \ If CNT < 33 then jump to PL37, as this is the right CPX #33 \ half of the circle and the sign of the x-coordinate is BCC PL37 \ correct EOR #%11111111 \ This is the left half of the circle, so we want to ADC #0 \ flip the sign of the x-coordinate in (T A) using two's TAX \ complement, so we start with the low byte and store it \ in X (the ADC adds 1 as we know the C flag is set) LDA #&FF \ And then we flip the high byte in T ADC #0 STA T TXA \ Finally, we restore the low byte from X, so we have \ now negated the x-coordinate in (T A) CLC \ Clear the C flag so we can do some more addition below .PL37 ADC K3 \ We now calculate the following: STA K6 \ \ K6(1 0) = (T A) + K3(1 0) \ \ to add the coordinates of the centre to our circle \ point, starting with the low bytes LDA K3+1 \ And then doing the high bytes, so we now have: ADC T \ STA K6+1 \ K6(1 0) = K * sin(CNT) + K3(1 0) \ \ which is the result we want for the x-coordinate LDA CNT \ Set A = CNT + 16 CLC ADC #16 JSR FMLTU2 \ Call FMLTU2 to calculate: \ \ A = K * sin(A) \ = K * sin(CNT + 16) \ = K * cos(CNT) TAX \ Set X = A \ = K * cos(CNT) LDA #0 \ Set T = 0, so we have the following: STA T \ \ (T X) = K * cos(CNT) \ \ which is the y-coordinate of the circle for this count LDA CNT \ Set A = (CNT + 15) mod 64 ADC #15 AND #63 CMP #33 \ If A < 33 (i.e. CNT is 0-16 or 48-64) then jump to BCC PL38 \ PL38, as this is the bottom half of the circle and the \ sign of the y-coordinate is correct TXA \ This is the top half of the circle, so we want to EOR #%11111111 \ flip the sign of the y-coordinate in (T X) using two's ADC #0 \ complement, so we start with the low byte in X (the TAX \ ADC adds 1 as we know the C flag is set) LDA #&FF \ And then we flip the high byte in T, so we have ADC #0 \ now negated the y-coordinate in (T X) STA T CLC \ Clear the C flag so we can do some more addition below .PL38 JSR BLINE \ Call BLINE to draw this segment, which also increases \ CNT by STP, the step size CMP #65 \ If CNT >= 65 then skip the next instruction BCS P%+5 JMP PLL3 \ Jump back for the next segment CLC \ Clear the C flag to indicate success RTS \ Return from the subroutine .CIRCLE2 \ This is the entry point for this subroutine STZ LSP \ Reset the ball line heap by setting the ball line heap \ pointer to 0 JSR CIRCLE3 \ Call CIRCLE3 to populate the ball line heap \ Fall through into LS2FL to send the ball line heap to \ the I/O processor for drawing on-screen