Question
In: Computer Science
4. Explain what is happening on each line of the following AVR assembly code. If you...
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
LDIR20, 0x74
LDIR21, 0x04
LDIR22, 0x22
ADDR20, R22
SUBR22, R21
ADDR20, R21
MOVR20, R21
JMPDONE
ADDR21, R20
SUBR21, R22
DONE:SUBR20, R21
-embedded system-
Solutions
Expert Solution
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
venereology answered 1 year agoRelated Solutions
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