.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 ;LDX #8 ; This instruction is commented out in the original ; source 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 BBC Micro 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 BBC Micro ; 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 big 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 [Show more] Type: Subroutine Category: Maths (Arithmetic) Summary: Calculate (A P) = P * X Deep dive: Shift-and-add multiplicationContext: See this subroutine in context in the source code References: This subroutine is called as follows: * SQUA2 calls MU11
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.