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Elite on the BBC Micro

Maths (Arithmetic): LL28 [Master version]

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 antilog,X \ table .LLfix STA R \ Set the result in R to the value of A RTS \ Return from the subroutine 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