Question

4. Explain what is happening on each line of the following AVR assembly code. If you were to execute this code what would be the final decimal values in R20, R21 and SREG registers?

BCLR 0

BCLR 1

BCLR 2

BCLR 3

BCLR 4

BCLR 5

BCLR 6

BCLR 7

LDI ​R19, 0x02

LDI​R20, 0x74

LDI​R21, 0x04

LDI​R22, 0x22

ADD​R20, R22

SUB​R22, R21

ADD​R20, R21

MOV​R20, R21

JMP​DONE

ADD​R21, R20

SUB​R21, R22

DONE:​SUB​R20, R21

-embedded system-

Answer 1

Answer

BCLR 0 => Clearing 0th bit in SREG register (carry flag)

BCLR 1 => Clearing 1st bit in SREG register (Zero flag)

BCLR 2 => Clearing 2nd bit in SREG register (Negative flag)

BCLR 3 => Clearing 3rd bit in SREG register (Overflow flag)

BCLR 4 => Clearing 4th bit in SREG register (Sign flag)

BCLR 5 => Clearing 5th bit in SREG register (Half carru flag)

BCLR 6 => Clearing 6th bit in SREG register (Copy storage flag)

BCLR 7 => Clearing 7th bit in SREG register (Disables interrupt)

LDI ​R19, 0x02 => Loads value 0x02 in R19 => R19 = 0x02

LDI ​R20, 0x74 => Loads value 0x74 in R20 => R20 = 0x74

LDI​ R21, 0x04 => Loads value 0x04 in R21 => R21 = 0x04

LDI​ R22, 0x22 => Loads value 0x22 in R22 => R22 = 0x22

ADD​ R20, R22 => Add R20 and R22 and Store it into R20 => R20 = 0x74+0x22 = 0x96

SUB​ R22, R21 => Subtract R21 from R22 and store it into R22 => R22 = 0x22 - 0x04 = 0x1E

ADD​ R20, R21 => Add R20 and R21 and Store it into R20 => R20 = 0x96+0x04 = 0x9A

MOV ​R20, R21 => Move R21 value to R20 => R20 = 0x04

JMP​ DONE => Unconditional Jump

ADD ​R21, R20 => Skipped

SUB​ R21, R22 => Skipped

DONE:​ SUB​ R20, R21 => Subtract R21 from R20 and store it into R20 => R20 = 0x04 - 0x04 = 0x00

=> Enables Zero Flag in SREG register

Final Values :

R20 = 0x00
R21 = 0x04

SREG = 0x02