Q.Subtract 23H from 3FH when CY = 0, write the program and solve.

The 8051 has only SUBB (subtract with borrow), so the carry is first cleared with CLR C to make it behave as a plain subtraction; the accumulator holds the minuend and the source is subtracted from …

[!TLDR]

Clear the carry, load 3FH into A and SUBB 23H; A becomes 1CH and CY = 0 marks a positive result.

Concept - subtraction with SUBB. The 8051 provides only SUBB A, source (subtract with borrow), which computes A = A - source - CY. To perform an ordinary subtraction the carry must be cleared first (CLR C) so that no borrow is included. Internally the processor takes the 2's complement of the source, adds it to A and inverts the carry; after the operation, CY = 0 means the result is positive while CY = 1 means it is negative (and A then holds the 2's complement of the magnitude).

Program:

CLR C          ; make CY = 0
MOV A, #3FH    ; load 3FH into A
MOV R3, #23H   ; load 23H into R3
SUBB A, R3     ; A = A - R3, result in A

Working:

  A  ->  3FH     0011 1111              0011 1111
  R3 ->  23H   - 0010 0011   2's compl = + 1101 1101 …