Question

Analyze the following code and identify the final result stored in R20 after execution. Show your answer in Hexadecimal format.

LDI R20, 0xE5
SWAP R20
CLC
ROR R20
ROR R20

Answer 1

After executing the given instructions, the final result stored in R20 is 0x97 in hexadecimal format.

Step-by-step explanation:

The provided code is an assembly sequence operating on a microcontroller register (R20). Let's analyze each instruction one by one:

1. LDI R20, 0xE5: This loads the hexadecimal value 0xE5 into R20. The binary representation of 0xE5 is 1110 0101.
2. SWAP R20: This instruction swaps the nibbles of R20. Therefore, the value becomes 0101 1110 in binary, or 0x5E in hexadecimal.
3. CLC: This clears the carry flag. For the subsequent operations, this means that the carry-in for a rotate or shift operation is zero.
4. ROR R20: Rotate right through carry. Since the carry is cleared, the result after the first ROR operation will be 0010 1111 with the carry out being '0'. In hexadecimal, this is 0x2F.
5. ROR R20: Another rotate right through carry. The final result is 1001 0111 with the carry out being '0'. This is 0x97 in hexadecimal.

Thus, the final value stored in R20 after the execution of the code is 0x97.