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Elite on the BBC Micro

Maths (Arithmetic): FMLTU [Master version]

Name: FMLTU [Show more] Type: Subroutine Category: Maths (Arithmetic) Summary: Calculate A = A * Q / 256 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: * DOEXP calls FMLTU * LL51 calls FMLTU * LL9 (Part 5 of 12) calls FMLTU * MVEIT (Part 3 of 9) calls FMLTU * PLS22 calls FMLTU

Do the following multiplication of two unsigned 8-bit numbers, returning only the high byte of the result: (A ?) = A * Q or, to put it another way: A = A * Q / 256 The Master and 6502 Second Processor versions use logarithms to speed up the multiplication process. See the deep dive on "Multiplication using logarithms" for more details.
.FMLTU STX P \ Store X in P so we can preserve it through the call to \ FMULTU STA widget \ Store A in widget, so now widget = argument A TAX \ Transfer A into X, so now X = argument A BEQ MU3 \ If A = 0, jump to MU3 to return a result of 0, as \ 0 * Q / 256 is always 0 \ We now want to calculate La + Lq, 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 La \ = low byte of La (as we set X to A above) LDX Q \ Set X = Q BEQ MU3again \ If X = 0, jump to MU3again to return a result of 0, as \ A * 0 / 256 is always 0 CLC \ Set A = A + low byte of Lq ADC logL,X \ = low byte of La + low byte of Lq LDA log,X \ Set A = high byte of Lq LDX widget \ Set A = A + C + high byte of La ADC log,X \ = high byte of Lq + high byte of La + C \ \ so we now have: \ \ A = high byte of (La + Lq) BCC MU3again \ If the addition fitted into one byte and didn't carry, \ then La + Lq < 256, so we jump to MU3again to return a \ result of 0 TAX \ Otherwise La + Lq >= 256, so we return the A-th entry LDA antilog,X \ from the antilog table LDX P \ Restore X from P so it is preserved RTS \ Return from the subroutine .MU3again LDA #0 \ Set A = 0 .MU3 \ If we get here then A (our result) is already 0 LDX P \ Restore X from P so it is preserved RTS \ Return from the subroutine