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Maths (Arithmetic): MLS1

[BBC Master version]

Name: MLS1 [Show more] Type: Subroutine Category: Maths (Arithmetic) Summary: Calculate (A P) = ALP1 * A
Context: See this subroutine in context in the source code References: This subroutine is called as follows: * STARS1 calls MLS1 * STARS6 calls MLS1 * STARS1 calls via MULTS-2 * STARS2 calls via MULTS-2 * STARS6 calls via MULTS-2

Calculate the following: (A P) = ALP1 * A where ALP1 is the magnitude of the current roll angle alpha, in the range 0-31. This routine uses an unrolled version of MU11. MU11 calculates P * X, so we use the same algorithm but with P set to ALP1 and X set to A. The unrolled version here can skip the bit tests for bits 5-7 of P as we know P < 32, so only 5 shifts with bit tests are needed (for bits 0-4), while the other 3 shifts can be done without a test (for bits 5-7).
Other entry points: MULTS-2 Calculate (A P) = X * A
.MLS1 LDX ALP1 \ Set P to the roll angle alpha magnitude in ALP1 STX P \ (0-31), so now we calculate P * A .MULTS TAX \ Set X = A, so now we can calculate P * X instead of \ P * A to get our result, and we can use the algorithm \ from MU11 to do that, just unrolled (as MU11 returns \ P * X) AND #%10000000 \ Set T to the sign bit of A STA T TXA \ Set A = |A| AND #127 BEQ MU6 \ If A = 0, jump to MU6 to set P(1 0) = 0 and return \ from the subroutine using a tail call TAX \ Set T1 = X - 1 DEX \ STX T1 \ 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 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, but instead of using a \ loop like MU11, we just unroll it, starting with bit 0 BCC P%+4 \ If C (i.e. the next bit from P) is set, do the ADC T1 \ addition for this bit of P: \ \ A = A + T1 + 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 the shift-and-add loop for bit 1 ADC T1 ROR A ROR P BCC P%+4 \ Repeat the shift-and-add loop for bit 2 ADC T1 ROR A ROR P BCC P%+4 \ Repeat the shift-and-add loop for bit 3 ADC T1 ROR A ROR P BCC P%+4 \ Repeat the shift-and-add loop for bit 4 ADC T1 ROR A ROR P LSR A \ Just do the "shift" part for bit 5 ROR P LSR A \ Just do the "shift" part for bit 6 ROR P LSR A \ Just do the "shift" part for bit 7 ROR P ORA T \ Give A the sign bit of the original argument A that \ we put into T above RTS \ Return from the subroutine