- A
Prepend a NOP sled and inject shellcode into the buffer
Why wrong: ASLR randomizes memory regions, so the shellcode address is unpredictable.
- B
Return-to-libc (ret2libc)
Why wrong: ret2libc requires knowing the libc base, which is randomized by ASLR.
- C
Return to the PLT entry for system()
The PLT (Procedure Linkage Table) address is not randomized by ASLR, allowing reliable calls to library functions.
- D
Use a ROP chain to bypass ASLR and execute shellcode
Why wrong: ROP can bypass ASLR but is more complex; return-to-plt is more direct.
Quick Answer
The answer is to return to the PLT entry for system(). This technique is the most effective because Address Space Layout Randomization (ASLR) randomizes stack and library addresses, making direct shellcode injection unreliable, but the Procedure Linkage Table (PLT) within the binary itself retains a fixed, known address for dynamically linked functions like system(). By overwriting the return address with the PLT entry for system() and placing a pointer to a command string such as "/bin/sh" on the stack, the attacker achieves code execution without needing to know any runtime addresses, and the absence of stack canaries leaves the buffer overflow unguarded. On the CompTIA PenTest+ PT0-002 exam, this scenario tests your understanding of bypassing ASLR when stack protections are weak—a common trap is assuming shellcode is always viable, but the PLT provides a static target. Memory tip: "PLT is the fixed lifeline when ASLR randomizes the rest."
PT0-002 Attacks and Exploits Practice Question
This PT0-002 practice question tests your understanding of attacks and exploits. The scenario asks you to isolate a root cause — eliminate options that address a different problem before choosing. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.
A penetration tester is exploiting a Linux system that has ASLR enabled but no stack canaries. The vulnerability is a classic stack-based buffer overflow. Which of the following is the most effective method to achieve code execution?
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
Return to the PLT entry for system()
Option C is correct because with ASLR enabled, direct shellcode injection fails due to randomized stack addresses, but the PLT entry for system() has a fixed, known address in the binary. Since there are no stack canaries, a simple buffer overflow can overwrite the return address to jump directly to system() in the PLT, passing a pointer to a command string (e.g., "/bin/sh") already in memory, achieving code execution without needing to know runtime addresses.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
Prepend a NOP sled and inject shellcode into the buffer
Why it's wrong here
ASLR randomizes memory regions, so the shellcode address is unpredictable.
- ✗
Return-to-libc (ret2libc)
Why it's wrong here
ret2libc requires knowing the libc base, which is randomized by ASLR.
- ✓
Return to the PLT entry for system()
Why this is correct
The PLT (Procedure Linkage Table) address is not randomized by ASLR, allowing reliable calls to library functions.
Related concept
Read the scenario before looking for a memorised answer.
- ✗
Use a ROP chain to bypass ASLR and execute shellcode
Why it's wrong here
ROP can bypass ASLR but is more complex; return-to-plt is more direct.
Common exam traps
Common exam trap: answer the scenario, not the keyword
The trap here is that candidates assume ASLR always prevents code execution, but they overlook that PLT entries have fixed addresses in the binary, making ret2plt a viable bypass when no stack canaries are present.
Detailed technical explanation
How to think about this question
The PLT (Procedure Linkage Table) is part of the executable's .plt section, which contains stubs that jump to the GOT (Global Offset Table). For dynamically linked functions like system(), the PLT entry is resolved lazily by the dynamic linker, but its address within the binary is fixed and known at compile time. In practice, this technique works because the attacker can overwrite the return address with the PLT address of system() and arrange for the stack to contain a pointer to a string like "/bin/sh" (often found in libc or placed in an environment variable), enabling a shell without needing to execute arbitrary shellcode.
KKey Concepts to Remember
- Read the scenario before looking for a memorised answer.
- Find the constraint that changes the correct option.
- Eliminate answers that are true in general but not in this case.
TExam Day Tips
- Watch for words such as best, first, most likely and least administrative effort.
- Review why wrong options are wrong, not only why the correct option is correct.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
How this comes up in practice
A security team runs a vulnerability scan on a web application and discovers an unpatched SQL injection flaw. The team prioritises remediation by CVSS score — critical flaws are patched within 24 hours, high within 7 days. Questions like this test whether you understand vulnerability management processes, scanning tools, and remediation prioritisation.
What to study next
Got this wrong? Here's your next step.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
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FAQ
Questions learners often ask
What does this PT0-002 question test?
Attacks and Exploits — This question tests Attacks and Exploits — Read the scenario before looking for a memorised answer..
What is the correct answer to this question?
The correct answer is: Return to the PLT entry for system() — Option C is correct because with ASLR enabled, direct shellcode injection fails due to randomized stack addresses, but the PLT entry for system() has a fixed, known address in the binary. Since there are no stack canaries, a simple buffer overflow can overwrite the return address to jump directly to system() in the PLT, passing a pointer to a command string (e.g., "/bin/sh") already in memory, achieving code execution without needing to know runtime addresses.
What should I do if I get this PT0-002 question wrong?
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
What is the key concept behind this question?
Read the scenario before looking for a memorised answer.
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Last reviewed: Jun 25, 2026
This PT0-002 practice question is part of Courseiva's free CompTIA certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the PT0-002 exam.
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