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BBC Micro Elite

Universe: TT24

Name: TT24 [View in context] Type: Subroutine Category: Universe Summary: Calculate system data from the system seeds
Calculate system data from the seeds in QQ15 and store them in the relevant locations. Specifically, this routine calculates the following from the three 16-bit seeds in QQ15 (using only w0_hi, w1_hi and w1_lo): QQ3 = economy (0-7) QQ4 = government (0-7) QQ5 = technology level (0-14) QQ6 = population * 10 (1-71) QQ7 = productivity (96-62480) The ranges of the various values are shown in brackets. Note that the radius and type of inhabitant are calculated on-the-fly in the TT25 routine when the system data gets displayed, so they aren't calculated here.
.TT24 { LDA QQ15+1 \ Fetch w0_hi and extract bits 0-2 to determine the AND #%00000111 \ system's economy, and store in QQ3 STA QQ3 LDA QQ15+2 \ Fetch w1_lo and extract bits 3-5 to determine the LSR A \ system's government, and store in QQ4 LSR A LSR A AND #%00000111 STA QQ4 LSR A \ If government isn't anarchy or feudal, skip to TT77, BNE TT77 \ as we need to fix the economy of anarchy and feudal \ systems so they can't be rich LDA QQ3 \ Set bit 1 of the economy in QQ3 to fix the economy ORA #%00000010 \ for anarchy and feudal governments STA QQ3 .TT77 LDA QQ3 \ Now to work out the tech level, which we do like this: EOR #%00000111 \ CLC \ flipped_economy + (w1_hi AND %11) + (government / 2) STA QQ5 \ \ or, in terms of memory locations: \ \ QQ5 = (QQ3 EOR %111) + (QQ15+3 AND %11) + (QQ4 / 2) \ \ We start by setting QQ5 = QQ3 EOR %111 LDA QQ15+3 \ We then take the first 2 bits of w1_hi (QQ15+3) and AND #%00000011 \ add it into QQ5 ADC QQ5 STA QQ5 LDA QQ4 \ And finally we add QQ4 / 2 and store the result in LSR A \ QQ5, using LSR then ADC to divide by 2, which rounds ADC QQ5 \ up the result for odd-numbered government types STA QQ5 ASL A \ Now to work out the population, like so: ASL A \ ADC QQ3 \ (tech level * 4) + economy + government + 1 ADC QQ4 \ ADC #1 \ or, in terms of memory locations: STA QQ6 \ \ QQ6 = (QQ5 * 4) + QQ3 + QQ4 + 1 LDA QQ3 \ Finally, we work out productivity, like this: EOR #%00000111 \ ADC #3 \ (flipped_economy + 3) * (government + 4) STA P \ * population LDA QQ4 \ * 8 ADC #4 \ STA Q \ or, in terms of memory locations: JSR MULTU \ \ QQ7 = (QQ3 EOR %111 + 3) * (QQ4 + 4) * QQ6 * 8 \ \ We do the first step by setting P to the first \ expression in brackets and Q to the second, and \ calling MULTU, so now (A P) = P * Q. The highest this \ can be is 10 * 11 (as the maximum values of economy \ and government are 7), so the high byte of the result \ will always be 0, so we actually have: \ \ P = P * Q \ = (flipped_economy + 3) * (government + 4) LDA QQ6 \ We now take the result in P and multiply by the STA Q \ population to get the productivity, by setting Q to JSR MULTU \ the population from QQ6 and calling MULTU again, so \ now we have: \ \ (A P) = P * population ASL P \ Next we multiply the result by 8, as a 16-bit number, ROL A \ so we shift both bytes to the left three times, ASL P \ using the carry flag to carry bits from bit 7 of the ROL A \ low byte into bit 0 of the high byte ASL P ROL A STA QQ7+1 \ Finally, we store the productivity in two bytes, with LDA P \ the low byte in QQ7 and the high byte in QQ7+1 STA QQ7 RTS \ Return from the subroutine }