According to the memory view given below. If RO = 0x20000004, after the LDMIA rO, (r7, r3, r9} instruction is executed, which of the following is the r3 and r0 register contents, respectively (data layout to.g endian)?
According to the memory view given below. If RO = 0x20000004, after the LDMIA rO, (r7, r3, r9} instruction is executed, which of the following is the r3 and r0 register contents, respectively (data layout to.g endian)?
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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14--> ) According to the memory view given below. If RO = 0x20000004, after the LDMIA rO, (r7, r3, r9} instruction is executed, which of the following is the r3 and r0 register contents, respectively (data layout to.g endian)?

Transcribed Image Text:### Memory Layout and Register Initialization Example
#### Explanation
In the image, a memory layout is provided, and a task is given to execute a specific command in the context of register initialization. Here’s the text transcription and analysis of the depicted content.
#### Transcription
"Aşağıda verilen bellek görünümüne göre, R0 = 0x20000004 ise LDMIA r0, {r7, r3, r9} komutu çalıştırıldıktan sonra, r3 ve r0 register içerikleri sırası ile aşağıdakilerden hangisidir (veri yerleşimi big endian)?"
| Memory Address | Memory Content |
|:--------------:|:---------------:|
| 0x20000013 | 0x79 |
| 0x20000012 | 0xcd |
| 0x20000011 | 0xa3 |
| 0x20000010 | 0xfd |
| 0x2000000F | 0x12 |
| 0x2000000E | 0x0d |
| 0x2000000D | 0xe0 |
| 0x2000000C | 0xeb |
| 0x2000000B | 0x2c |
| 0x2000000A | 0x1a |
| 0x20000009 | 0xdc |
| 0x20000008 | 0x11 |
| 0x20000007 | 0x34 |
| 0x20000006 | 0xce |
| 0x20000005 | 0xfd |
| 0x20000004 | 0xe4 |
| 0x20000003 | 0xa2 |
| 0x20000002 | 0x3f |
| 0x20000001 | 0x06 |
| 0x20000000 | 0xf5 |
#### Detailed Explanation
- **Problem Context**:
Based on the given memory layout, with the starting address stored in register R0 set to 0x20000004, the command `LDMIA r0, {r7, r
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