Moving: MVT1

       Name: MVT1                                                    [Show more]
       Type: Subroutine
   Category: Moving
    Summary: Calculate (x_sign x_hi x_lo) = (x_sign x_hi x_lo) + (A R)
    Context: See this subroutine in context in the source code
 References: This subroutine is called as follows:
             * MVEIT (Part 6 of 9) calls MVT1
             * SFS2 calls MVT1
             * MVEIT (Part 3 of 9) calls via MVT1-2


 Add the signed delta (A R) to a ship's coordinate, along the axis given in X.
 Mathematically speaking, this routine translates the ship along a single axis
 by a signed delta. Taking the example of X = 0, the x-axis, it does the
 following:

   (x_sign x_hi x_lo) = (x_sign x_hi x_lo) + (A R)

 (In practice, MVT1 is only ever called directly with A = 0 or 128, otherwise
 it is always called via MVT-2, which clears A apart from the sign bit. The
 routine is written to cope with a non-zero delta_hi, so it supports a full
 16-bit delta, but it appears that delta_hi is only ever used to hold the
 sign of the delta.)

 The comments below assume we are adding delta to the x-axis, though the axis
 is determined by the value of X.


 Arguments:

   (A R)                The signed delta, so A = delta_hi and R = delta_lo

   X                    Determines which coordinate axis of INWK to change:

                          * X = 0 adds the delta to (x_lo, x_hi, x_sign)

                          * X = 3 adds the delta to (y_lo, y_hi, y_sign)

                          * X = 6 adds the delta to (z_lo, z_hi, z_sign)


 Other entry points:

   MVT1-2               Clear bits 0-6 of A before entering MVT1


 AND #%10000000         \ Clear bits 0-6 of A

.MVT1

 ASL A                  \ Set the C flag to the sign bit of the delta, leaving
                        \ delta_hi << 1 in A

 STA S                  \ Set S = delta_hi << 1
                        \
                        \ This also clears bit 0 of S

 LDA #0                 \ Set T = just the sign bit of delta (in bit 7)
 ROR A
 STA T

 LSR S                  \ Set S = delta_hi >> 1
                        \       = |delta_hi|
                        \
                        \ This also clear the C flag, as we know that bit 0 of
                        \ S was clear before the LSR

 EOR INWK+2,X           \ If T EOR x_sign has bit 7 set, then x_sign and delta
 BMI MV10               \ have different signs, so jump to MV10

                        \ At this point, we know x_sign and delta have the same
                        \ sign, that sign is in T, and S contains |delta_hi|,
                        \ so now we want to do:
                        \
                        \   (x_sign x_hi x_lo) = (x_sign x_hi x_lo) + (S R)
                        \
                        \ and then set the sign of the result to the same sign
                        \ as x_sign and delta

 LDA R                  \ First we add the low bytes, so:
 ADC INWK,X             \
 STA INWK,X             \   x_lo = x_lo + R

 LDA S                  \ Then we add the high bytes:
 ADC INWK+1,X           \
 STA INWK+1,X           \   x_hi = x_hi + S

 LDA INWK+2,X           \ And finally we add any carry into x_sign, and if the
 ADC #0                 \ sign of x_sign and delta in T is negative, make sure
 ORA T                  \ the result is negative (by OR'ing with T)
 STA INWK+2,X

 RTS                    \ Return from the subroutine

.MV10

                        \ If we get here, we know x_sign and delta have
                        \ different signs, with delta's sign in T, and
                        \ |delta_hi| in S, so now we want to do:
                        \
                        \   (x_sign x_hi x_lo) = (x_sign x_hi x_lo) - (S R)
                        \
                        \ and then set the sign of the result according to
                        \ the signs of x_sign and delta

 LDA INWK,X             \ First we subtract the low bytes, so:
 SEC                    \
 SBC R                  \   x_lo = x_lo - R
 STA INWK,X

 LDA INWK+1,X           \ Then we subtract the high bytes:
 SBC S                  \
 STA INWK+1,X           \   x_hi = x_hi - S

 LDA INWK+2,X           \ And finally we subtract any borrow from bits 0-6 of
 AND #%01111111         \ x_sign, and give the result the opposite sign bit to T
 SBC #0                 \ (i.e. give it the sign of the original x_sign)
 ORA #%10000000
 EOR T
 STA INWK+2,X

 BCS MV11               \ If the C flag is set by the above SBC, then our sum
                        \ above didn't underflow and is correct - to put it
                        \ another way, (x_sign x_hi x_lo) >= (S R) so the result
                        \ should indeed have the same sign as x_sign, so jump to
                        \ MV11 to return from the subroutine

                        \ Otherwise our subtraction underflowed because
                        \ (x_sign x_hi x_lo) < (S R), so we now need to flip the
                        \ subtraction around by using two's complement to this:
                        \
                        \   (S R) - (x_sign x_hi x_lo)
                        \
                        \ and then we need to give the result the same sign as
                        \ (S R), the delta, as that's the dominant figure in the
                        \ sum

 LDA #1                 \ First we subtract the low bytes, so:
 SBC INWK,X             \
 STA INWK,X             \   x_lo = 1 - x_lo

 LDA #0                 \ Then we subtract the high bytes:
 SBC INWK+1,X           \
 STA INWK+1,X           \   x_hi = 0 - x_hi

 LDA #0                 \ And then we subtract the sign bytes:
 SBC INWK+2,X           \
                        \   x_sign = 0 - x_sign

 AND #%01111111         \ Finally, we set the sign bit to the sign in T, the
 ORA T                  \ sign of the original delta, as the delta is the
 STA INWK+2,X           \ dominant figure in the sum

.MV11

 RTS                    \ Return from the subroutine