Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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![### Program Execution and Memory Analysis
#### 1. Program Execution Starting at Memory Address 100
**Memory Table:**
| Address | Contents |
|---------|----------|
| 100 | 5800 |
| 101 | 1801 |
| 102 | 6800 |
| ... | ... |
| 800 | 001F |
| 801 | 000A |
| 802 | 000C |
**Question 1:**
Upon the completion of the program, what will be the contents of the following memory address?
- Memory address 800 contents: [____]
---
#### 2. Program Execution Starting at Memory Address 210
**Memory Table:**
| Address | Contents |
|---------|----------|
| 210 | 5801 |
| 211 | 2802 |
| 212 | 6801 |
| ... | ... |
| 800 | FFFF |
| 801 | 000A |
| 802 | 0005 |
**Question 2:**
Upon the completion of the program, what will be the contents of the following memory address?
- Memory address 801 contents: [____]
- Answer all questions with hexadecimal.
---
These exercises involve understanding how a program interacts with memory and how operations alter memory addresses. Remember to analyze how each instruction affects the given memory contents.](https://content.bartleby.com/qna-images/question/44c3b237-e4ca-4e38-a278-251c2495f1b6/611c06d8-c7b9-4884-8ba2-3ad1617936b3/7nq3mq_processed.png)
Transcribed Image Text:### Program Execution and Memory Analysis
#### 1. Program Execution Starting at Memory Address 100
**Memory Table:**
| Address | Contents |
|---------|----------|
| 100 | 5800 |
| 101 | 1801 |
| 102 | 6800 |
| ... | ... |
| 800 | 001F |
| 801 | 000A |
| 802 | 000C |
**Question 1:**
Upon the completion of the program, what will be the contents of the following memory address?
- Memory address 800 contents: [____]
---
#### 2. Program Execution Starting at Memory Address 210
**Memory Table:**
| Address | Contents |
|---------|----------|
| 210 | 5801 |
| 211 | 2802 |
| 212 | 6801 |
| ... | ... |
| 800 | FFFF |
| 801 | 000A |
| 802 | 0005 |
**Question 2:**
Upon the completion of the program, what will be the contents of the following memory address?
- Memory address 801 contents: [____]
- Answer all questions with hexadecimal.
---
These exercises involve understanding how a program interacts with memory and how operations alter memory addresses. Remember to analyze how each instruction affects the given memory contents.
Transcribed Image Text:### Instruction Format and Opcodes
In this section, we illustrate the instruction format and provide a list of opcodes used in a hypothetical computing system. This foundational understanding is essential for decoding and executing machine-level instructions.
#### Instruction Format
The instruction format is divided into two parts:
- **Opcode (15..12)**: This segment of the instruction specifies the operation to be performed. It consists of the four most significant bits.
- **Address (11..0)**: The remaining bits define the memory address, which is used in conjunction with the opcode to execute the instruction.
> *Note that all memory contents and addresses are hexadecimal.*
#### A Portion of the Instruction Set
The following table presents a portion of the instruction set, detailing various opcodes and their associated operations:
| Opcode | Instruction |
|--------|------------------------------------------|
| 0101 | Load AC from memory |
| 0110 | Store AC to memory |
| 0001 | Add AC from memory - AC + memory |
| 0010 | Subtract memory from AC - AC - memory |
| 0011 | Multiply AC by memory - AC * memory |
| 0100 | Divide AC by memory - AC / memory |
These opcodes demonstrate basic arithmetic and data transfer operations using an accumulator (AC) and memory interaction. This set is crucial for programmers working on low-level system programming to understand the manipulation and processing of data at the hardware level.
Expert Solution
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Step 1
Solution 1-
Upon completion of program the value of 800 content inside the memory location is 0028H.
- For First content 5800 the opcode is 5 and adress is 800 .Now converting opcode 5 to binary we get 0101, which is given in table as load accumulator in memory .The initial 101 instructions load the memory address 800 to AC.
- For second content 1801 the opcode is 1 so instruction by converting 1 to binary we get 0001 which is the second instructions that adds the contents at memory location 801 to AC.
- For 6800 , the third instruction Stores current value of AC to memory location.
- So value inside 800 is result of addition of contents of 800 and 801 which can be seen by proper solution of given program.
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