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BBC Micro Elite

Moving: MVT3

Name: MVT3 [View in context] Type: Subroutine Category: Moving Summary: Calculate K(3 2 1) = (x_sign x_hi x_lo) + K(3 2 1)
Add an INWK position coordinate - i.e. x, y or z - to K(3 2 1), like this: K(3 2 1) = (x_sign x_hi x_lo) + K(3 2 1) The INWK coordinate to add to K(3 2 1) is specified by X. Arguments: X The coordinate to add to K(3 2 1), as follows: * If X = 0, add (x_sign x_hi x_lo) * If X = 3, add (y_sign y_hi y_lo) * If X = 6, add (z_sign z_hi z_lo) Returns: A Contains a copy of the high byte of the result, K+3 X X is preserved
.MVT3 LDA K+3 \ Set S = K+3 STA S AND #%10000000 \ Set T = sign bit of K(3 2 1) STA T EOR INWK+2,X \ If x_sign has a different sign to K(3 2 1), jump to BMI MV13 \ MV13 to process the addition as a subtraction LDA K+1 \ Set K(3 2 1) = K(3 2 1) + (x_sign x_hi x_lo) CLC \ starting with the low bytes ADC INWK,X STA K+1 LDA K+2 \ Then the middle bytes ADC INWK+1,X STA K+2 LDA K+3 \ And finally the high bytes ADC INWK+2,X AND #%01111111 \ Setting the sign bit of K+3 to T, the original sign ORA T \ of K(3 2 1) STA K+3 RTS \ Return from the subroutine .MV13 LDA S \ Set S = |K+3| (i.e. K+3 with the sign bit cleared) AND #%01111111 STA S LDA INWK,X \ Set K(3 2 1) = (x_sign x_hi x_lo) - K(3 2 1) SEC \ starting with the low bytes SBC K+1 STA K+1 LDA INWK+1,X \ Then the middle bytes SBC K+2 STA K+2 LDA INWK+2,X \ And finally the high bytes, doing A = |x_sign| - |K+3| AND #%01111111 \ and setting the C flag for testing below SBC S ORA #%10000000 \ Set the sign bit of K+3 to the opposite sign of T, EOR T \ i.e. the opposite sign to the original K(3 2 1) STA K+3 BCS MV14 \ If the C flag is set, i.e. |x_sign| >= |K+3|, then \ the sign of K(3 2 1). In this case, we want the \ result to have the same sign as the largest argument, \ which is (x_sign x_hi x_lo), which we know has the \ opposite sign to K(3 2 1), and that's what we just set \ the sign of K(3 2 1) to... so we can jump to MV14 to \ return from the subroutine LDA #1 \ We need to swap the sign of the result in K(3 2 1), SBC K+1 \ which we do by calculating 0 - K(3 2 1), which we can STA K+1 \ do with 1 - C - K(3 2 1), as we know the C flag is \ clear. We start with the low bytes LDA #0 \ Then the middle bytes SBC K+2 STA K+2 LDA #0 \ And finally the high bytes SBC K+3 AND #%01111111 \ Set the sign bit of K+3 to the same sign as T, ORA T \ i.e. the same sign as the original K(3 2 1), as STA K+3 \ that's the largest argument .MV14 RTS \ Return from the subroutine