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Suppose the following code fragment is executed and that the data word that is allocated and initialized to OxEFABCD89 is located in memory at address 4000. Also assume that we are using Big Endian byte ordering. .data M: .word OxEFABCD89 .text addi $1, $0, M Ibu $2, 2($1) lb $3,3($1)

Suppose the following code fragment is executed and that the data word that is allocated and initialized to OxEFABCD89 is located in memory at address 4000. Also assume that we are using Big Endian byte ordering. .data M: .word OxEFABCD89 .text addi $1, $0, M Ibu $2, 2($1) lb $3,3($1)

Computer Networking: A Top-Down Approach (7th Edition)
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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### Byte Ordering and Assembly Code Analysis #### Problem Statement: Suppose the following code fragment is executed and that the data word that is allocated and initialized to `0xEFABCD89` is located in memory at address 4000. Also, assume that we are using Big Endian byte ordering. ```assembly .data M: .word 0xEFABCD89 .text addi $1, $0, M lbu $2, 2($1) lb $3, 3($1) ``` #### Explanation: Let's break down the code to understand what happens step-by-step. 1. **Data Segment Initialization:** - `M` is a label pointing to a word initialized to `0xEFABCD89`. - In Big Endian, the bytes are ordered from the most significant byte to the least significant byte. So, memory at address 4000 contains: ``` Address: 4000 4001 4002 4003 Value: 0xEF 0xAB 0xCD 0x89 ``` 2. **Text Segment Instructions:** - `addi $1, $0, M` - This means: `$1 = $0 + M` - Register `$0` is always `0`, so `$1` will hold the address of `M`, which is `4000`. - Thus, `$1` = 4000. - `lbu $2, 2($1)` - Load the byte located at offset 2 from the address in `$1` into `$2`, unsigned. - `$1` holds 4000, so we load the byte at address `4002`. - At `4002`, the byte is `0xCD`. - Since it's unsigned, `$2 = 0x000000CD`. - `lb $3, 3($1)` - Load the byte located at offset 3 from the address in `$1` into `$3`, signed. - `$1` holds 4000, so we load the byte at address `4003`. - At `4003`, the byte is `0x89`, which is negative when interpreted as signed (because the most significant bit is 1). - Using two's complement, `0x
Transcribed Image Text:### Byte Ordering and Assembly Code Analysis #### Problem Statement: Suppose the following code fragment is executed and that the data word that is allocated and initialized to `0xEFABCD89` is located in memory at address 4000. Also, assume that we are using Big Endian byte ordering. ```assembly .data M: .word 0xEFABCD89 .text addi $1, $0, M lbu $2, 2($1) lb $3, 3($1) ``` #### Explanation: Let's break down the code to understand what happens step-by-step. 1. **Data Segment Initialization:** - `M` is a label pointing to a word initialized to `0xEFABCD89`. - In Big Endian, the bytes are ordered from the most significant byte to the least significant byte. So, memory at address 4000 contains: ``` Address: 4000 4001 4002 4003 Value: 0xEF 0xAB 0xCD 0x89 ``` 2. **Text Segment Instructions:** - `addi $1, $0, M` - This means: `$1 = $0 + M` - Register `$0` is always `0`, so `$1` will hold the address of `M`, which is `4000`. - Thus, `$1` = 4000. - `lbu $2, 2($1)` - Load the byte located at offset 2 from the address in `$1` into `$2`, unsigned. - `$1` holds 4000, so we load the byte at address `4002`. - At `4002`, the byte is `0xCD`. - Since it's unsigned, `$2 = 0x000000CD`. - `lb $3, 3($1)` - Load the byte located at offset 3 from the address in `$1` into `$3`, signed. - `$1` holds 4000, so we load the byte at address `4003`. - At `4003`, the byte is `0x89`, which is negative when interpreted as signed (because the most significant bit is 1). - Using two's complement, `0x
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