Q.Analyse the following program.

CLR C  
MOV A, #5CH    ; load A with value 5CH (A = 5CH)  
SUBB A, #7EH   ; subtract 7EH from A  
INC NEXT       ; if CY = 0, jump to NEXT target  
CPL A          ; if CY = 1, then take 1's compliment  
INC A          ; and increment to get 2's compliment  
NEXT: MOV R1, A ; save A in R1  

Because the subtrahend 7EH is larger than the minuend 5CH, SUBB produces a borrow, setting CY = 1 and leaving A holding the 2's-complement form of the result; taking the 2's complement back recovers the magnitude. …

[!TLDR]

SUBB of 7EH from 5CH borrows, so CY = 1 (negative); complementing and incrementing A recovers the magnitude 22H, i.e. the result is -22H.

Concept - sign of a SUBB result. After SUBB A, source, CY = 0 means the result is positive and already correct, while CY = 1 means the result is negative and A holds the 2's complement of the magnitude. The 2's complement is undone by taking the 1's complement (CPL A) and then incrementing (INC A). (The line commented "if CY = 0, jump to NEXT target" is intended as a jump-if-no-carry to the NEXT label.)

Trace:

• CLR C - clear the carry so SUBB is a plain subtract.
• MOV A, #5CH - A = 5CH.
• SUBB A, #7EH - compute 5CH - 7EH. Since 5CH (92) < 7EH (126), a borrow occurs: CY = 1 and A holds the 2's complement of the magnitude.

  A  ->  5CH     0101 1100                0101 1100
         7EH   - 0111 1110   2's compl = + 1000 0010
              -----------                -----------
        -22H                           0 1101 1110 …