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Elite on the BBC Micro

Version analysis of LL120

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

Code variations between these versions are shown below.

Name: LL120 Type: Subroutine Category: Maths (Arithmetic) Summary: Calculate (Y X) = (S x1_lo) * XX12+2 or (S x1_lo) / XX12+2
Calculate the following: * If T = 0 (more vertical than horizontal), (Y X) = (S x1_lo) * XX12+2 * If T <> 0 (more horizontal than vertical), (Y X) = (S x1_lo) / XX12+2 giving (Y X) the opposite sign to the slope direction in XX12+3. Other entry points: LL122 Calculate (Y X) = (S R) * Q and set the sign to the opposite of the top byte on the stack
.LL120 LDA XX15 \ Set R = x1_lo STA R

Code variation 1 of 1A variation in the comments only

This variation is blank in the Disc (flight), Disc (docked), 6502 Second Processor, Master and Electron versions.

Cassette

\.LL120 \ This label is commented out in the original source
 JSR LL129              \ Call LL129 to do the following:
                        \
                        \   Q = XX12+2
                        \     = line gradient
                        \
                        \   A = S EOR XX12+3
                        \     = S EOR slope direction
                        \
                        \   (S R) = |S R|
                        \
                        \ So A contains the sign of S * slope direction

 PHA                    \ Store A on the stack so we can use it later

 LDX T                  \ If T is non-zero, so it's more horizontal than
 BNE LL121              \ vertical, jump down to LL121 to calculate this
                        \ instead:
                        \
                        \   (Y X) = (S R) / Q

.LL122

                        \ The following calculates:
                        \
                        \   (Y X) = (S R) * Q
                        \
                        \ using the same shift-and-add algorithm that's
                        \ documented in MULT1

 LDA #0                 \ Set A = 0

 TAX                    \ Set (Y X) = 0 so we can start building the answer here
 TAY

 LSR S                  \ Shift (S R) to the right, so we extract bit 0 of (S R)
 ROR R                  \ into the C flag

 ASL Q                  \ Shift Q to the left, catching bit 7 in the C flag

 BCC LL126              \ If C (i.e. the next bit from Q) is clear, do not do
                        \ the addition for this bit of Q, and instead skip to
                        \ LL126 to just do the shifts

.LL125

 TXA                    \ Set (Y X) = (Y X) + (S R)
 CLC                    \
 ADC R                  \ starting with the low bytes
 TAX

 TYA                    \ And then doing the high bytes
 ADC S
 TAY

.LL126

 LSR S                  \ Shift (S R) to the right
 ROR R

 ASL Q                  \ Shift Q to the left, catching bit 7 in the C flag

 BCS LL125              \ If C (i.e. the next bit from Q) is set, loop back to
                        \ LL125 to do the addition for this bit of Q

 BNE LL126              \ If Q has not yet run out of set bits, loop back to
                        \ LL126 to do the "shift" part of shift-and-add until
                        \ we have done additions for all the set bits in Q, to
                        \ give us our multiplication result

 PLA                    \ Restore A, which we calculated above, from the stack

 BPL LL133              \ If A is positive jump to LL133 to negate (Y X) and
                        \ return from the subroutine using a tail call

 RTS                    \ Return from the subroutine

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