.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 \ Set up a counter in X to count the 8 bits in P LSR P \ Set P = P >> 1 \ and C flag = bit 0 of P \ We are now going to work our way through the bits of \ P, and do a shift-add for any bits that are set, \ keeping the running total in A. We just did the first \ shift right, so we now need to do the first add and \ loop through the other bits in P .MUL6 BCC P%+4 \ If C (i.e. the next bit from 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 DEX \ Decrement the loop counter BNE MUL6 \ Loop back for the next bit until P has been rotated \ all the way 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.
Label MUL6 is local to this routine