WREG contains the value 0x86. What will be the content of the WREG (in hex notation) and the states of the status bits Z and C after executing the following instruction? addlw 0x89 WREG= [Type your answer here.] Z= [Type your answer here.] C= [Type your answer here.]

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**Problem Statement**

WREG contains the value 0x86. What will be the content of the WREG (in hex notation) and the states of the status bits Z and C after executing the following instruction?

    addlw 0x89

**Solution Fields**

- WREG = [Type your answer here.]
- Z = [Type your answer here.]
- C = [Type your answer here.]

---

**Instruction Explanation**

- **addlw 0x89**: This instruction adds the literal value `0x89` to the contents of the WREG register.

**Status Bits**

- **Z (Zero bit)**: This bit is set if the result of the addition is zero.
- **C (Carry bit)**: This bit is set if there is a carry out from the most significant bit during the addition.

When executing the `addlw` instruction, calculate the result by adding `0x86` and `0x89`:

1. **Perform the Addition**: 
    - 0x86 (134 in decimal)
    - 0x89 (137 in decimal)
    - Result: 0x86 + 0x89 = 0x11F (287 in decimal)

2. **Determine the WREG Content**:
    - Since the result 0x11F is beyond 8-bit storage, only the lower 8 bits are stored in WREG, giving 0x1F.

3. **Status Bit Evaluation**:
    - **Z** is 0 because the result (0x1F) is not zero.
    - **C** is 1 because there is a carry out from the most significant bit (as the sum exceeds 255).

---

**Resulting Values**

- WREG = 0x1F
- Z = 0
- C = 1

---
Transcribed Image Text:--- **Problem Statement** WREG contains the value 0x86. What will be the content of the WREG (in hex notation) and the states of the status bits Z and C after executing the following instruction? addlw 0x89 **Solution Fields** - WREG = [Type your answer here.] - Z = [Type your answer here.] - C = [Type your answer here.] --- **Instruction Explanation** - **addlw 0x89**: This instruction adds the literal value `0x89` to the contents of the WREG register. **Status Bits** - **Z (Zero bit)**: This bit is set if the result of the addition is zero. - **C (Carry bit)**: This bit is set if there is a carry out from the most significant bit during the addition. When executing the `addlw` instruction, calculate the result by adding `0x86` and `0x89`: 1. **Perform the Addition**: - 0x86 (134 in decimal) - 0x89 (137 in decimal) - Result: 0x86 + 0x89 = 0x11F (287 in decimal) 2. **Determine the WREG Content**: - Since the result 0x11F is beyond 8-bit storage, only the lower 8 bits are stored in WREG, giving 0x1F. 3. **Status Bit Evaluation**: - **Z** is 0 because the result (0x1F) is not zero. - **C** is 1 because there is a carry out from the most significant bit (as the sum exceeds 255). --- **Resulting Values** - WREG = 0x1F - Z = 0 - C = 1 ---
**Exercise Instructions:**

Please answer the previous question again, this time assuming that variables A and B contain the values 0xFF and 0x80, respectively.

- WREG = [Type your answer here.]
- A = [Type your answer here.]
- B = [Type your answer here.]
- Z = [Type your answer here.]
- C = [Type your answer here.] 

**Note to Students:**

Make sure to provide your answers based on the given hexadecimal values. Consider the roles of these variables within the context of the problem you are solving, and ensure your calculations are accurate. If you need to review related concepts, please refer to the resources provided in earlier lessons.
Transcribed Image Text:**Exercise Instructions:** Please answer the previous question again, this time assuming that variables A and B contain the values 0xFF and 0x80, respectively. - WREG = [Type your answer here.] - A = [Type your answer here.] - B = [Type your answer here.] - Z = [Type your answer here.] - C = [Type your answer here.] **Note to Students:** Make sure to provide your answers based on the given hexadecimal values. Consider the roles of these variables within the context of the problem you are solving, and ensure your calculations are accurate. If you need to review related concepts, please refer to the resources provided in earlier lessons.
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