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Elite on the BBC Micro

Ship hanger: HANGER [Master version]

Name: HANGER [Show more] Type: Subroutine Category: Ship hanger Summary: Display the ship hanger
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: * HALL calls HANGER * HAS2 calls entry point HA3 * HAS3 calls entry point HA3

This routine is called after the ships in the hanger have been drawn, so all it has to do is draw the hanger's background. The hanger background is made up of two parts: * The hanger floor consists of 11 screen-wide horizontal lines, which start out quite spaced out near the bottom of the screen, and bunch ever closer together as the eye moves up towards the horizon, where they merge to give a sense of perspective * The back wall of the hanger consists of 15 equally spaced vertical lines that join the horizon to the top of the screen The ships in the hangar have already been drawn by this point, so the lines are drawn so they don't overlap anything that's already there, which makes them look like they are behind and below the ships. This is achieved by drawing the lines in from the screen edges until they bump into something already on-screen. For the horizontal lines, when there are multiple ships in the hanger, this also means drawing lines between the ships, as well as in from each side. Other entry points: HA3 Contains an RTS
.HANGER \ We start by drawing the floor LDX #2 \ We start with a loop using a counter in T that goes \ from 2 to 12, one for each of the 11 horizontal lines \ in the floor, so set the initial value in X LDA #%00001111 \ Set bits 1 and 2 of the Access Control Register at STA VIA+&34 \ SHEILA &34 to switch screen memory into &3000-&7FFF .HAL1 STX T \ Store the loop counter in T LDA #130 \ Set A = 130 STX Q \ Set Q = T JSR DVID4_DUPLICATE \ Calculate the following: \ \ (P R) = 256 * A / Q \ = 256 * 130 / T \ \ so P = 130 / T, and as the counter T goes from 2 to \ 12, P goes 65, 43, 32 ... 13, 11, 10, with the \ difference between two consecutive numbers getting \ smaller as P gets smaller \ \ We can use this value as a y-coordinate to draw a set \ of horizontal lines, spaced out near the bottom of the \ screen (high value of P, high y-coordinate, lower down \ the screen) and bunching up towards the horizon (low \ value of P, low y-coordinate, higher up the screen) LDA P \ Set Y = #Y + P CLC \ ADC #Y \ where #Y is the y-coordinate of the centre of the TAY \ screen, so Y is now the horizontal pixel row of the \ line we want to draw to display the hanger floor LDA ylookup,Y \ Look up the page number of the character row that STA SC+1 \ contains the pixel with the y-coordinate in Y, and \ store it in the high byte of SC(1 0) at SC+1 STA R \ Also store the page number in R LDA P \ Set the low byte of SC(1 0) to the y-coordinate mod 7, AND #7 \ which determines the pixel row in the character block STA SC \ we need to draw in (as each character row is 8 pixels \ high), so SC(1 0) now points to the address of the \ start of the horizontal line we want to draw LDY #0 \ Set Y = 0 so the call to HAS2 starts drawing the line \ in the first byte of the screen row, at the left edge \ of the screen JSR HAS2 \ Draw a horizontal line from the left edge of the \ screen, going right until we bump into something \ already on-screen, at which point stop drawing LDY R \ Fetch the page number of the line from R, increment it INY \ so it points to the right half of the character row STY SC+1 \ (as each row takes up 2 pages), and store it in the \ high byte of SC(1 0) at SC+1 LDA #%01000000 \ Now to draw the same line but from the right edge of \ the screen, so set a pixel mask in A to check the \ second pixel of the last byte, so we skip the 2-pixel \ scren border at the right edge of the screen LDY #248 \ Set Y = 248 so the call to HAS3 starts drawing the \ line in the last byte of the screen row, at the right \ edge of the screen JSR HAS3 \ Draw a horizontal line from the right edge of the \ screen, going left until we bump into something \ already on-screen, at which point stop drawing LDY HCNT \ Fetch the value of HCNT, which gets set to 0 in the \ HALL routine above if there is only one ship BEQ HA2 \ If HCNT is zero, jump to HA2 to skip the following \ as there is only one ship in the hanger \ If we get here then there are multiple ships in the \ hanger, so we also need to draw the horizontal line in \ the gap between the ships LDY #0 \ First we draw the line from the centre of the screen \ to the right. SC(1 0) points to the start address of \ the second half of the screen row, so we set Y to 0 so \ the call to HAL3 starts drawing from the first \ character in that second half LDA #%10001000 \ We want to start drawing from the first pixel, so we \ set a mask in A to the first pixel in the 4-pixel byte JSR HAL3 \ Call HAL3, which draws a line from the halfway point \ across the right half of the screen, going right until \ we bump into something already on-screen, at which \ point it stops drawing DEC SC+1 \ Decrement the high byte of SC(1 0) in SC+1 to point to \ the previous page (i.e. the left half of this screen \ row) LDY #248 \ We now draw the line from the centre of the screen \ to the left. SC(1 0) points to the start address of \ the first half of the screen row, so we set Y to 248 \ so the call to HAS3 starts drawing from the last \ character in that first half LDA #%00010000 \ We want to start drawing from the last pixel, so we \ set a mask in A to the last pixel in the 4-pixel byte JSR HAS3 \ Call HAS3, which draws a line from the halfway point \ across the left half of the screen, going left until \ we bump into something already on-screen, at which \ point it stops drawing .HA2 \ We have finished threading our horizontal line behind \ the ships already on-screen, so now for the next line LDX T \ Fetch the loop counter from T and increment it INX CPX #13 \ If the loop counter is less than 13 (i.e. T = 2 to 12) BCC HAL1 \ then loop back to HAL1 to draw the next line \ The floor is done, so now we move on to the back wall LDA #60 \ Set S = 60, so we run the following 60 times (though I STA S \ have no idea why it's 60 times, when it should be 15, \ as this has the effect of drawing each vertical line \ four times, each time starting one character row lower \ on-screen) LDA #16 \ We want to draw 15 vertical lines, one every 16 pixels \ across the screen, with the first at x-coordinate 16, \ so set this in A to act as the x-coordinate of each \ line as we work our way through them from left to \ right, incrementing by 16 for each new line LDX #&40 \ Set X = &40, the high byte of the start of screen STX R \ memory (the screen starts at location &4000) and the \ page number of the first screen row .HAL6 LDX R \ Set the high byte of SC(1 0) to R STX SC+1 STA T \ Store A in T so we can retrieve it later AND #%11111100 \ A contains the x-coordinate of the line to draw, and STA SC \ each character block is 4 pixels wide, so setting the \ low byte of SC(1 0) to A mod 4 points SC(1 0) to the \ correct character block on the top screen row for this \ x-coordinate LDX #%10001000 \ Set a mask in X to the first pixel in the 4-pixel byte LDY #1 \ We are going to start drawing the line from the second \ pixel from the top (to avoid drawing on the 1-pixel \ border), so set Y to 1 to point to the second row in \ the first character block .HAL7 TXA \ Copy the pixel mask to A AND (SC),Y \ If the pixel we want to draw is non-zero (using A as a BNE HA6 \ mask), then this means it already contains something, \ so jump to HA6 to stop drawing this line TXA \ Copy the pixel mask to A again AND #RED \ Apply the pixel mask in A to a four-pixel block of \ red pixels, so we now know which bits to set in screen \ memory ORA (SC),Y \ OR the byte with the current contents of screen \ memory, so the pixel we want is set to red (because \ we know the bits are already 0 from the above test) STA (SC),Y \ Store the updated pixel in screen memory INY \ Increment Y to point to the next row in the character \ block, i.e. the next pixel down CPY #8 \ Loop back to HAL7 to draw this next pixel until we BNE HAL7 \ have drawn all 8 in the character block INC SC+1 \ There are two pages of memory for each character row, INC SC+1 \ so we increment the high byte of SC(1 0) twice to \ point to the same character but in the next row down LDY #0 \ Set Y = 0 to point to the first row in this character \ block BEQ HAL7 \ Loop back up to HAL7 to keep drawing the line (this \ BEQ is effectively a JMP as Y is always zero) .HA6 LDA T \ Fetch the x-coordinate of the line we just drew from T CLC \ into A, and add 16 so that A contains the x-coordinate ADC #16 \ of the next line to draw BCC P%+4 \ If the addition overflowed, increment the page number INC R \ in R to point to the second half of the screen row DEC S \ Decrement the loop counter in S BNE HAL6 \ Loop back to HAL6 until we have run through the loop \ 60 times, by which point we are most definitely done IF _SNG47 LDA #%00001001 \ Clear bits 1 and 2 of the Access Control Register at STA VIA+&34 \ SHEILA &34 to switch main memory back into &3000-&7FFF RTS \ Return from the subroutine (this instruction is not \ needed as we could just fall through into the RTS at \ HA3 below) ELIF _COMPACT JMP BULB2 \ Jump to BULB2 to switch main memory back into \ &3000-&7FFF and return from the subroutine ENDIF .HA3 RTS \ Return from the subroutine