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BBC Micro Elite

Maths (Arithmetic): ADD

Name: ADD [View in context] Type: Subroutine Category: Maths (Arithmetic) Summary: Calculate (A X) = (A P) + (S R) Deep dive: Adding sign-magnitude numbers
Add two 16-bit sign-magnitude numbers together, calculating: (A X) = (A P) + (S R)
.ADD STA T1 \ Store argument A in T1 AND #%10000000 \ Extract the sign (bit 7) of A and store it in T STA T EOR S \ EOR bit 7 of A with S. If they have different bit 7s BMI MU8 \ (i.e. they have different signs) then bit 7 in the \ EOR result will be 1, which means the EOR result is \ negative. So the AND, EOR and BMI together mean "jump \ to MU8 if A and S have different signs" \ If we reach here, then A and S have the same sign, so \ we can add them and set the sign to get the result LDA R \ Add the least significant bytes together into X: CLC \ ADC P \ X = P + R TAX LDA S \ Add the most significant bytes together into A. We ADC T1 \ stored the original argument A in T1 earlier, so we \ can do this with: \ \ A = A + S + C \ = T1 + S + C ORA T \ If argument A was negative (and therefore S was also \ negative) then make sure result A is negative by \ OR-ing the result with the sign bit from argument A \ (which we stored in T) RTS \ Return from the subroutine .MU8 \ If we reach here, then A and S have different signs, \ so we can subtract their absolute values and set the \ sign to get the result LDA S \ Clear the sign (bit 7) in S and store the result in AND #%01111111 \ U, so U now contains |S| STA U LDA P \ Subtract the least significant bytes into X: SEC \ SBC R \ X = P - R TAX LDA T1 \ Restore the A of the argument (A P) from T1 and AND #%01111111 \ clear the sign (bit 7), so A now contains |A| SBC U \ Set A = |A| - |S| \ At this point we have |A P| - |S R| in (A X), so we \ need to check whether the subtraction above was the \ the right way round (i.e. that we subtracted the \ smaller absolute value from the larger absolute \ value) BCS MU9 \ If |A| >= |S|, our subtraction was the right way \ round, so jump to MU9 to set the sign \ If we get here, then |A| < |S|, so our subtraction \ above was the wrong way round (we actually subtracted \ the larger absolute value from the smaller absolute \ value). So let's subtract the result we have in (A X) \ from zero, so that the subtraction is the right way \ round STA U \ Store A in U TXA \ Set X = 0 - X using two's complement (to negate a EOR #&FF \ number in two's complement, you can invert the bits ADC #1 \ and add one - and we know the C flag is clear as we TAX \ didn't take the BCS branch above, so the ADC will do \ the correct addition) LDA #0 \ Set A = 0 - A, which we can do this time using a SBC U \ a subtraction with the C flag clear ORA #%10000000 \ We now set the sign bit of A, so that the EOR on the \ next line will give the result the opposite sign to \ argument A (as T contains the sign bit of argument \ A). This is the same as giving the result the same \ sign as argument S (as A and S have different signs), \ which is what we want, as S has the larger absolute \ value .MU9 EOR T \ If we get here from the BCS above, then |A| >= |S|, \ so we want to give the result the same sign as \ argument A, so if argument A was negative, we flip \ the sign of the result with an EOR (to make it \ negative) RTS \ Return from the subroutine