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BBC Micro Elite

Moving: MVT6

Name: MVT6 [View in context] Type: Subroutine Category: Moving Summary: Calculate (A P+2 P+1) = (x_sign x_hi x_lo) + (A P+2 P+1)
Do the following calculation, for the coordinate given by X (so this is what it does for the x-coordinate): (A P+2 P+1) = (x_sign x_hi x_lo) + (A P+2 P+1) A is a sign bit and is not included in the calculation, but bits 0-6 of A are preserved. Bit 7 is set to the sign of the result. Arguments: A The sign of P(2 1) in bit 7 P(2 1) The 16-bit value we want to add the coordinate to X The coordinate to add, as follows: * If X = 0, add to (x_sign x_hi x_lo) * If X = 3, add to (y_sign y_hi y_lo) * If X = 6, add to (z_sign z_hi z_lo) Returns: A The sign of the result (in bit 7)
.MVT6 TAY \ Store argument A into Y, for later use EOR INWK+2,X \ Set A = A EOR x_sign BMI MV50 \ If the sign is negative, i.e. A and x_sign have \ different signs, jump to MV50 \ The signs are the same, so we can add the two \ arguments and keep the sign to get the result LDA P+1 \ First we add the low bytes: CLC \ ADC INWK,X \ P+1 = P+1 + x_lo STA P+1 LDA P+2 \ And then the high bytes: ADC INWK+1,X \ STA P+2 \ P+2 = P+2 + x_hi TYA \ Restore the original A argument that we stored earlier \ so that we keep the original sign RTS \ Return from the subroutine .MV50 LDA INWK,X \ First we subtract the low bytes: SEC \ SBC P+1 \ P+1 = x_lo - P+1 STA P+1 LDA INWK+1,X \ And then the high bytes: SBC P+2 \ STA P+2 \ P+2 = x_hi - P+2 BCC MV51 \ If the last subtraction underflowed, then the C flag \ will be clear and x_hi < P+2, so jump to MV51 to \ negate the result TYA \ Restore the original A argument that we stored earlier EOR #%10000000 \ but flip bit 7, which flips the sign. We do this \ because x_hi >= P+2 so we want the result to have the \ same sign as x_hi (as it's the dominant side in this \ calculation). The sign of x_hi is x_sign, and x_sign \ has the opposite sign to A, so we flip the sign in A \ to return the correct result RTS \ Return from the subroutine .MV51 LDA #1 \ Our subtraction underflowed, so we negate the result SBC P+1 \ using two's complement, first with the low byte: STA P+1 \ \ P+1 = 1 - P+1 LDA #0 \ And then the high byte: SBC P+2 \ STA P+2 \ P+2 = 0 - P+2 TYA \ Restore the original A argument that we stored earlier \ as this is the correct sign for the result. This is \ because x_hi < P+2, so we want to return the same sign \ as P+2, the dominant side RTS \ Return from the subroutine