.LCASH STX T1 \ Subtract the least significant bytes: LDA CASH+3 \ SEC \ CASH+3 = CASH+3 - X SBC T1 STA CASH+3 STY T1 \ Then the second most significant bytes: LDA CASH+2 \ SBC T1 \ CASH+2 = CASH+2 - Y STA CASH+2 LDA CASH+1 \ Then the third most significant bytes (which are 0): SBC #0 \ STA CASH+1 \ CASH+1 = CASH+1 - 0 LDA CASH \ And finally the most significant bytes (which are 0): SBC #0 \ STA CASH \ CASH = CASH - 0 BCS TT113 \ If the C flag is set then the subtraction didn't \ underflow, so the value in CASH is correct and we can \ jump to TT113 to return from the subroutine with the \ C flag set to indicate success (as TT113 contains an \ RTS) \ Otherwise we didn't have enough cash in CASH to \ subtract (Y X) from it, so fall through into \ MCASH to reverse the sum and restore the original \ value in CASH, and returning with the C flag clearName: LCASH [Show more] Type: Subroutine Category: Maths (Arithmetic) Summary: Subtract an amount of cash from the cash potContext: See this subroutine in context in the source code References: This subroutine is called as follows: * TT219 calls LCASH * eq calls LCASH

Subtract (Y X) cash from the cash pot in CASH, but only if there is enough cash in the pot. As CASH is a four-byte number, this calculates: CASH(0 1 2 3) = CASH(0 1 2 3) - (0 0 Y X) Returns: C flag If set, there was enough cash to do the subtraction If clear, there was not enough cash to do the subtraction

[X]

Entry point TT113 in subroutine MCASH (category: Maths (Arithmetic))

Contains an RTS