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Mastering Malware Analysis

You're reading from   Mastering Malware Analysis A malware analyst's practical guide to combating malicious software, APT, cybercrime, and IoT attacks

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Product type Paperback
Published in Sep 2022
Publisher Packt
ISBN-13 9781803240244
Length 572 pages
Edition 2nd Edition
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Authors (2):
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Amr Thabet Amr Thabet
Author Profile Icon Amr Thabet
Amr Thabet
Alexey Kleymenov Alexey Kleymenov
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Alexey Kleymenov
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Toc

Table of Contents (20) Chapters Close

Preface 1. Part 1 Fundamental Theory
2. Chapter 1: Cybercrime, APT Attacks, and Research Strategies FREE CHAPTER 3. Chapter 2: A Crash Course in Assembly and Programming Basics 4. Part 2 Diving Deep into Windows Malware
5. Chapter 3: Basic Static and Dynamic Analysis for x86/x64 6. Chapter 4: Unpacking, Decryption, and Deobfuscation 7. Chapter 5: Inspecting Process Injection and API Hooking 8. Chapter 6: Bypassing Anti-Reverse Engineering Techniques 9. Chapter 7: Understanding Kernel-Mode Rootkits 10. Part 3 Examining Cross-Platform and Bytecode-Based Malware
11. Chapter 8: Handling Exploits and Shellcode 12. Chapter 9: Reversing Bytecode Languages – .NET, Java, and More 13. Chapter 10: Scripts and Macros – Reversing, Deobfuscation, and Debugging 14. Part 4 Looking into IoT and Other Platforms
15. Chapter 11: Dissecting Linux and IoT Malware 16. Chapter 12: Introduction to macOS and iOS Threats 17. Chapter 13: Analyzing Android Malware Samples 18. Index 19. Other Books You May Enjoy

Moving from assembly to high-level programming languages

Developers mostly don’t write in assembly. Instead, they write in higher-level languages, such as C or C++, and the compiler converts this high-level code into a low-level representation in assembly language. In this section, we will look at different code blocks represented in the assembly.

Arithmetic statements

Let’s look at different C statements and how they are represented in the assembly. We will use Intel IA-32 for this example. The same concept applies to other assembly languages as well:

  • X = 50 (assuming 0x00010000 is the address of the X variable in memory):
    mov eax, 50
    mov dword ptr [00010000h], eax
  • X = Y + 50 (assuming 0x00010000 represents X and 0x00020000 represents Y):
    mov eax, dword ptr [00020000h]
    add eax, 50
    mov dword ptr [00010000h], eax
  • X = Y + (50 * 2):
    mov eax, dword ptr [00020000h]
    push eax    ; save Y for now
    mov eax, 50 ; do the multiplication first...
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