.MU11 DEX \ Set T = X - 1 STX T \ \ We subtract 1 as the C flag will be set when we want \ to do an addition in the loop below LDA #0 \ Set A = 0 so we can start building the answer in A TAX \ Copy A into X. There is a comment in the original \ source here that says "just in case", which refers to \ the MU11 routine in the cassette and disc versions, \ which set X to 0 (as they use X as a loop counter). \ The version here doesn't use a loop, but this \ instruction makes sure the unrolled version returns \ the same results as the loop versions, just in case \ something out there relies on MU11 returning X = 0 LSR P \ Set P = P >> 1 \ and C flag = bit 0 of P \ We now repeat the following four instruction block \ eight times, one for each bit in P. In the cassette \ and disc versions of Elite the following is done with \ a loop, but it is marginally faster to unroll the loop \ and have eight copies of the code, though it does take \ up a bit more memory (though that isn't a concern when \ you have a 6502 Second Processor) BCC P%+4 \ If C (i.e. bit 0 of P) is set, do the ADC T \ addition for this bit of P: \ \ A = A + T + C \ = A + X - 1 + 1 \ = A + X ROR A \ Shift A right to catch the next digit of our result, \ which the next ROR sticks into the left end of P while \ also extracting the next bit of P ROR P \ Add the overspill from shifting A to the right onto \ the start of P, and shift P right to fetch the next \ bit for the calculation into the C flag BCC P%+4 \ Repeat for the second time ADC T ROR A ROR P BCC P%+4 \ Repeat for the third time ADC T ROR A ROR P BCC P%+4 \ Repeat for the fourth time ADC T ROR A ROR P BCC P%+4 \ Repeat for the fifth time ADC T ROR A ROR P BCC P%+4 \ Repeat for the sixth time ADC T ROR A ROR P BCC P%+4 \ Repeat for the seventh time ADC T ROR A ROR P BCC P%+4 \ Repeat for the eighth time ADC T ROR A ROR P RTS \ Return from the subroutineName: MU11 [View in context] Type: Subroutine Category: Maths (Arithmetic) Summary: Calculate (A P) = P * X
Do the following multiplication of two unsigned 8-bit numbers: (A P) = P * X This uses the same shift-and-add approach as MULT1, but it's simpler as we are dealing with unsigned numbers in P and X. See the deep dive on "Shift-and-add multiplication" for a discussion of how this algorithm works.