Maths (Arithmetic): LL28

       Name: LL28                                                    [Show more]
       Type: Subroutine
   Category: Maths (Arithmetic)
    Summary: Calculate R = 256 * A / Q
  Deep dive: Multiplication and division using logarithms
    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:
             * ARCTAN calls LL28
             * LL145 (Part 3 of 4) calls LL28
             * LL61 calls LL28
             * LL9 (Part 3 of 12) calls LL28
             * LL9 (Part 8 of 12) calls LL28


 Calculate the following, where A < Q:

   R = 256 * A / Q

 This is a sister routine to LL61, which does the division when A >= Q.

 If A >= Q then 255 is returned and the C flag is set to indicate an overflow
 (the C flag is clear if the division was a success).

 The result is returned in one byte as the result of the division multiplied
 by 256, so we can return fractional results using integers.

 This routine uses the same logarithm algorithm that's documented in FMLTU,
 except it subtracts the logarithm values, to do a division instead of a
 multiplication.


 Returns:

   C flag               Set if the answer is too big for one byte, clear if the
                        division was a success


 Other entry points:

   LL28+4               Skips the A >= Q check and always returns with C flag
                        cleared, so this can be called if we know the division
                        will work

   LL31                 Skips the A >= Q check and does not set the R counter,
                        so this can be used for jumping straight into the
                        division loop if R is already set to 254 and we know the
                        division will work


.LL28

 CMP Q                  \ If A >= Q, then the answer will not fit in one byte,
 BCS LL2                \ so jump to LL2 to return 255

 STA widget             \ Store A in widget, so now widget = argument A

 TAX                    \ Transfer A into X, so now X = argument A

 BEQ LLfix              \ If A = 0, jump to LLfix to return a result of 0, as
                        \ 0 * Q / 256 is always 0

                        \ We now want to calculate log(A) - log(Q), first adding
                        \ the low bytes (from the logL table), and then the high
                        \ bytes (from the log table)

 LDA logL,X             \ Set A = low byte of log(X)
                        \       = low byte of log(A) (as we set X to A above)

 LDX Q                  \ Set X = Q

 SEC                    \ Set A = A - low byte of log(Q)
 SBC logL,X             \       = low byte of log(A) - low byte of log(Q)

 LDX widget             \ Set A = high byte of log(A) - high byte of log(Q)
 LDA log,X
 LDX Q
 SBC log,X

 BCS LL2                \ If the subtraction fitted into one byte and didn't
                        \ underflow, then log(A) - log(Q) < 256, so we jump to
                        \ LL2 return a result of 255

 TAX                    \ Otherwise we return the A-th entry from the antilog
 LDA alogh,X            \ table

.LLfix

 STA R                  \ Set the result in R to the value of A

 RTS                    \ Return from the subroutine

\.LL28                  \ These instructions are commented out in the original
\CMP Q                  \ source

 BCS LL2                \ If the subtraction fitted into one byte and didn't
                        \ underflow, then log(A) - log(Q) < 256, so we jump to
                        \ LL2 to return a result of 255

 LDX #254               \ Otherwise set the result in R to 254
 STX R

.LL31

 ASL A                  \ Shift A to the left

 BCS LL29               \ If bit 7 of A was set, then jump straight to the
                        \ subtraction

 CMP Q                  \ If A < Q, skip the following subtraction
 BCC P%+4

 SBC Q                  \ A >= Q, so set A = A - Q

 ROL R                  \ Rotate the counter in R 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)

 BCS LL31               \ If we still have set bits in R, loop back to LL31 to
                        \ do the next iteration of 7

 RTS                    \ R left with remainder of division

.LL29

 SBC Q                  \ A >= Q, so set A = A - Q

 SEC                    \ Set the C flag to rotate into the result in R

 ROL R                  \ Rotate the counter in R 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)

 BCS LL31               \ If we still have set bits in R, loop back to LL31 to
                        \ do the next iteration of 7

 LDA R                  \ Set A to the remainder in R

 RTS                    \ Return from the subroutine with R containing the
                        \ remainder of the division

.LL2

 LDA #255               \ The division is very close to 1, so return the closest
 STA R                  \ possible answer to 256, i.e. R = 255

 RTS                    \ Return from the subroutine