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- ; 8.2 A+B
- ; Eftersom vi, jævnfør den udleverede UCF fil, antager at BTN2 = BTNL og BTN3 = BTNU,
- ; antager vi også at R7= X"FC" istedet for X"FD".
- R0 <- M[R6] ; BTN 3/U, R6= X"FE" = "1111 1110" = MR6
- R1 <- M[R7] ; BTN 2/L, R7= X"FC" = "1111 1100" = MR4
- M[R4] <- R0 + R1 ; R4= X"F8", R5=X"F9"
- M[R5] <- X"00"
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