What value is in $2 after this instruction is executed: addiu $2, $0, -1 O $2 holds 1 = 0x00000001 O $2 holds -1 = 0xFFFFFFFF
What value is in $2 after this instruction is executed: addiu $2, $0, -1 O $2 holds 1 = 0x00000001 O $2 holds -1 = 0xFFFFFFFF
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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![### MIPS Instruction Analysis
#### Question:
What value is in `$2` after this instruction is executed:
```
addiu $2, $0, -1
```
#### Answer Choices:
- \( \circ \) `$2` holds 1 = 0x00000001
- \( \circ \) `$2` holds -1 = 0xFFFFFFFF
#### Explanation:
The instruction `addiu $2, $0, -1` is an immediate addition unsigned instruction in MIPS assembly language. It performs the addition of the immediate value `-1` to the contents of register `$0` and stores the result in register `$2`.
In this case, register `$0` in MIPS assembly is a fixed register that always holds the value `0`. The operation, therefore, is:
```
$2 = $0 + (-1)
```
Since `$0` is always `0`, we have:
```
$2 = 0 + (-1) = -1
```
In a 32-bit unsigned representation, `-1` is equivalent to `0xFFFFFFFF`. Therefore, the correct answer is:
\( \circ \) `$2` holds -1 = 0xFFFFFFFF](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37ffeb4c-90e5-4cb9-a6b6-d76475059d8f%2Fd01a7beb-6d6c-458c-8bbb-4fe0c8752ae8%2F2t3w4r5_processed.png&w=3840&q=75)
Transcribed Image Text:### MIPS Instruction Analysis
#### Question:
What value is in `$2` after this instruction is executed:
```
addiu $2, $0, -1
```
#### Answer Choices:
- \( \circ \) `$2` holds 1 = 0x00000001
- \( \circ \) `$2` holds -1 = 0xFFFFFFFF
#### Explanation:
The instruction `addiu $2, $0, -1` is an immediate addition unsigned instruction in MIPS assembly language. It performs the addition of the immediate value `-1` to the contents of register `$0` and stores the result in register `$2`.
In this case, register `$0` in MIPS assembly is a fixed register that always holds the value `0`. The operation, therefore, is:
```
$2 = $0 + (-1)
```
Since `$0` is always `0`, we have:
```
$2 = 0 + (-1) = -1
```
In a 32-bit unsigned representation, `-1` is equivalent to `0xFFFFFFFF`. Therefore, the correct answer is:
\( \circ \) `$2` holds -1 = 0xFFFFFFFF
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