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⌂ ∞ Elite on the BBC Micro and NES

Maths (Arithmetic): ADD

[NES version, Bank 7]

       Name: ADD                                                     [Show more]
       Type: Subroutine
   Category: Maths (Arithmetic)
    Summary: Calculate (A X) = (A P) + (S R)
  Deep dive: Adding sign-magnitude numbers

    Context: See this subroutine in context in the source code
 References: This subroutine is called as follows:
             * InSystemJump calls ADD
             * MVS5 calls ADD
             * PLS22 calls ADD
             * STARS1 calls ADD
             * STARS2 calls ADD
             * STARS6 calls ADD




 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
                        ; 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                  ; 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