Flag Register

Program Status Word (PSW) Register

The program status word (PSW) register, also referred to as the flag register, is an 8 bit register.

Only 6 bits are used

• These four are CY (carry), AC (auxiliary carry), P (parity), and OV (overflow)
• They are called conditional flags, meaning that they indicate some conditions that resulted after an instruction was executed
• The PSW3 and PSW4 are designed as RS0 and RS1, and are used to change the bank
• The two unused bits are user-definable

CY

AC

F0

RS1

RS0

OV

--

P

CY	PSW.7	Carry flag
AC	PSW.6	Auxiliary carry flag
--	PSW.5	Available to the user for general purpose
RS1	PSW.4	Register Bank selector bit 1
RS0	PSW.3	Register Bank selector bit 0
OV	PSW.2	Overflow flag
--	PSW.1	User definable bit
P	PSW.0	Parity flag. Set/cleared by hardware each instruction cycle to indicate an  odd/even number of 1 bits in the accumulator.

Instructions that affect flag bits

Instruction	CY	OV	AC
ADD	X	X	X
ADDC	X	X	X
SUBB	X	X	X
MUL	0	X
DIV	0	X
DA	X
RPC	X
PLC	X
SETB C	1
CLR C	0
CPL C	X
ANL C, bit	X
ANL C, /bit	X
ORL C, bit	X
ORL C, /bit	X
MOV C, bit	X
CJNE	X

Example:

Show the status of the CY, AC and P flag after the addition of 38H and 2FH in the following instructions.

                MOV A, #38H

                ADD A, #2FH      ;after the addition A=67H, CY=0

Solution:

                38           00111000

            + 2F            00101111

               67            01100111

CY = 0 since there is no carry beyond the D7 bit

AC = 1 since there is a carry from the D3 to the D4 bi

P = 1 since the accumulator has an odd number of 1s (it has five 1s)

Example:

Show the status of the CY, AC and P flag after the addition of 9CH and 64H in the following instructions.

MOV A, #9CH

ADD A, #64H      ;after the addition A=00H, CY=1

Solution:

9C           10011100

         + 64            01100100

         100             00000000

CY = 1 since there is a carry beyond the D7 bit

AC = 1 since there is a carry from the D3 to the D4 bi

P = 0 since the accumulator has an even number of 1s (it has zero 1s)

Example:

Show the status of the CY, AC and P flag after the addition of 88H and 93H in the following instructions.

MOV A, #88H

ADD A, #93H      ;after the addition A=1BH, CY=1

Solution:

88           10001000

         + 93           10010011

          11B           00011011

CY = 1 since there is a carry beyond the D7 bit

AC = 0 since there is no carry from the D3 to the D4 bi

P = 0 since the accumulator has an even number of 1s (it has four 1s)