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BBC Micro Elite

Maths (Arithmetic): MULT3

Name: MULT3 [View in context] Type: Subroutine Category: Maths (Arithmetic) Summary: Calculate K(3 2 1 0) = (A P+1 P) * Q
Calculate the following multiplication between a signed 24-bit number and a signed 8-bit number, returning the result as a signed 32-bit number: K(3 2 1 0) = (A P+1 P) * Q The algorithm is the same shift-and-add algorithm as in routine MULT1, but extended to cope with more bits. Returns: C flag The C flag is cleared
.MULT3 STA R \ Store the high byte of (A P+1 P) in R AND #%01111111 \ Set K+2 to |A|, the high byte of K(2 1 0) STA K+2 LDA Q \ Set A to bits 0-6 of Q, so A = |Q| AND #%01111111 BEQ MU5 \ If |Q| = 0, jump to MU5 to set K(3 2 1 0) to 0, \ returning from the subroutine with a tail call SEC \ Set T = |Q| - 1 SBC #1 STA T \ We now use the same shift-and-add algorithm as MULT1 \ to calculate the following: \ \ K(2 1 0) = K(2 1 0) * |Q| \ \ so we start with the first shift right, in which we \ take (K+2 P+1 P) and shift it right, storing the \ result in K(2 1 0), ready for the multiplication loop \ (so the multiplication loop actually calculates \ (|A| P+1 P) * |Q|, as the following sets K(2 1 0) to \ (|A| P+1 P) shifted right) LDA P+1 \ Set A = P+1 LSR K+2 \ Shift the high byte in K+2 to the right ROR A \ Shift the middle byte in A to the right and store in STA K+1 \ K+1 (so K+1 contains P+1 shifted right) LDA P \ Shift the middle byte in P to the right and store in ROR A \ K, so K(2 1 0) now contains (|A| P+1 P) shifted right STA K \ We now use the same shift-and-add algorithm as MULT1 \ to calculate the following: \ \ K(2 1 0) = K(2 1 0) * |Q| LDA #0 \ Set A = 0 so we can start building the answer in A LDX #24 \ Set up a counter in X to count the 24 bits in K(2 1 0) .MUL2 BCC P%+4 \ If C (i.e. the next bit from K) is set, do the ADC T \ addition for this bit of K: \ \ A = A + T + C \ = A + |Q| - 1 + 1 \ = A + |Q| ROR A \ Shift A right by one place to catch the next digit ROR K+2 \ next digit of our result in the left end of K(2 1 0), ROR K+1 \ while also shifting K(2 1 0) right to fetch the next ROR K \ bit for the calculation into the C flag \ \ On the last iteration of this loop, the bit falling \ off the end of K will be bit 0 of the original A, as \ we did one shift before the loop and we are doing 24 \ iterations. We set A to 0 before looping, so this \ means the loop exits with the C flag clear DEX \ Decrement the loop counter BNE MUL2 \ Loop back for the next bit until K(2 1 0) has been \ rotated all the way \ The result (|A| P+1 P) * |Q| is now in (A K+2 K+1 K), \ but it is positive and doesn't have the correct sign \ of the final result yet STA T \ Save the high byte of the result into T LDA R \ Fetch the sign byte from the original (A P+1 P) \ argument that we stored in R EOR Q \ EOR with Q so the sign bit is the same as that of \ (A P+1 P) * Q AND #%10000000 \ Extract the sign bit ORA T \ Apply this to the high byte of the result in T, so \ that A now has the correct sign for the result, and \ (A K+2 K+1 K) therefore contains the correctly signed \ result STA K+3 \ Store A in K+3, so K(3 2 1 0) now contains the result RTS \ Return from the subroutine