CT4 Solve

The memory banks of a microcontroller is given below. What will be the final values of the registers A & R1, after the execution of the codes?

MOV A, #07H
MOV 04H, 08H
MOV R1, #03H
ADDC A, R1
SWAP R1
CLR C
ORL A, 0AH
XCH A, 0EH
SETB C
ADD A, R1

Solution: State of Flag register (i.e. PSW),

CY	AC	FO	RS1	RS0	OV	-	P
1	0	0	1	0	0	X	1

Thus, current value of

• Carry Flag = 1 [CY]
• Memory Bank = $B_2$[RS1 AND RS0]

Note

Memory bank is selected using RS1 and RS0 bit. For details, see Question 6.a from MMAL Final Solve.

Memory addresses of:

• $B_0\to 00H-07H$
• $B_1\to 08H-0FH$
• $B_2\to 10H-17H$
• $B_3\to 18H-1FH$

We need those addresses during the execution of code.

1. MOV A, #07H $\to$ Load $07H$ in register A.
   A = $07H$
2. MOV 04H, 08H $\to$ Copy the value of memory bank address 08H to address 04H. Currently 08H $(B_1)$ holds value $05H$, which will be copied into address 04H $(B_0)$.
3. MOV R1, #03H $\to$ Load $03H$ in R1 register of current selected bank $B_2$. The address of R1 is 10H.
   R1 = $03H$
4. ADDC A, R1 $\to$ Add A with R1 and carry CY.
   A = A+R1+CY
A= $07H+03H+1$
   A = $0BH$
5. SWAP R1 $\to$ Swap the lower nibble with higher nibble of R1. The value of R1 becomes $30H$ from $03H$.
   R1 = $30H$
6. CLR C $\to$ Clear the value of carry bit.
   CY = $0$
7. ORL A, 0AH $\to$ Logical OR between register A and memory bank address0AH. Currently A= $0BH$ and 0AH = $01H$.
   A = $0BH\vee 01H$
   A = $0BH$
8. XCH A, 0EH $\to$ Exchange the value between register A and memory bank address 0EH. Currently A= $0BH$ and 0EH = $01H$.
   A = $01H$
   0EH = $0BH$
9. SETB C $\to$ Set bit of carry flag.
   CY = $1$
10. ADD A, R1 $\to$ Add A with R1.
    A = $01H+30H$
    A = $31H$

Thus the final values, A = $31H$, R1 = $30H$