.DVIDT STA P+1 \ Set P+1 = A, so P(1 0) = (A P) EOR Q \ Set T = the sign bit of A EOR Q, so it's 1 if A and Q AND #%10000000 \ have different signs, i.e. it's the sign of the result STA T \ of A / Q LDA #0 \ Set A = 0 for us to build a result LDX #16 \ Set a counter in X to count the 16 bits in P(1 0) ASL P \ Shift P(1 0) left ROL P+1 ASL Q \ Clear the sign bit of Q the C flag at the same time LSR Q .DVL2 ROL A \ Shift A to the left CMP Q \ If A < Q skip the following subtraction BCC P%+4 SBC Q \ Set A = A - Q \ \ Going into this subtraction we know the C flag is \ set as we passed through the BCC above, and we also \ know that A >= Q, so the C flag will still be set once \ we are done ROL P \ Rotate P(1 0) to the left, and catch the result bit ROL P+1 \ into the C flag (which will be a 0 if we didn't \ do the subtraction, or 1 if we did) DEX \ Decrement the loop counter BNE DVL2 \ Loop back for the next bit until we have done all 16 \ bits of P(1 0) LDA P \ Set A = P so the low byte is in the result in A ORA T \ Set A to the correct sign bit that we set in T above RTS \ Return from the subroutineName: DVIDT [Show more] Type: Subroutine Category: Maths (Arithmetic) Summary: Calculate (P+1 A) = (A P) / QContext: See this subroutine in context in the source code References: No direct references to this subroutine in this source file
Calculate the following integer division between sign-magnitude numbers: (P+1 A) = (A P) / Q This uses the same shift-and-subtract algorithm as TIS2.
[X]
Label DVL2 is local to this routine