Question 507 of 509

Attacks and Exploits →mediumMultiple ChoiceObjective-mapped

Quick Answer

The answer is Return-Oriented Programming (ROP) to call mprotect and then execute shellcode on the stack. This technique succeeds because the binary is statically linked, meaning no libc is available for a traditional ret2libc attack, and the non-executable stack (NX) prevents direct shellcode execution. By chaining small instruction sequences called gadgets found within the binary itself, the attacker can call mprotect() to mark the stack region as executable, then pivot execution to shellcode placed there. Since ASLR is enabled but the binary is not PIE, the code base address remains fixed, making ROP gadgets reliably addressable. On the CompTIA PenTest+ PT0-002 exam, this scenario tests your understanding of bypassing modern memory protections in a constrained environment—a common trap is assuming ret2libc works when the binary is statically linked. Memory tip: “Static binary, no libc to borrow; ROP the gadgets to mprotect tomorrow.”

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 attempting to exploit a Linux system that has ASLR and DEP enabled. The tester has identified a buffer overflow vulnerability in a network service compiled without stack canaries and with a non-executable stack (NX). The binary is statically linked and not PIE. Which exploitation technique is most likely to succeed under these conditions?

Clue words in this question

Noticing these words before you look at the options changes how you read each choice.

Clue: "most likely"
Why it matters: Probability qualifier — the question wants the most probable cause or outcome, not a guaranteed one. Eliminate low-probability options.

Question 1mediummultiple choice

Full question →

A
Heap spraying to place shellcode in the heap and then overwrite a function pointer to execute the shellcode
Why wrong: Heap spraying is often used when ASLR is present, but NX still prevents execution from the heap unless the heap is made executable. Also, the binary is not PIE, so addresses are known, but the heap may still be non-executable.
B
Return-to-libc attack using libc functions
Why wrong: Return-to-libc is a technique against NX, but since the binary is statically linked (no libc loaded dynamically), the attacker would need to use gadgets from the binary itself, not libc.
C
Return-Oriented Programming (ROP) to call mprotect and then execute shellcode on the stack
ROP allows the attacker to chain gadgets to call mprotect and change memory permissions on the stack to executable, then jump to shellcode placed on the stack. This bypasses NX while leveraging the known addresses from the statically linked, non-PIE binary.
D
Ret2plt to call system() via the PLT
Why wrong: Ret2plt is used against dynamically linked binaries to call functions from libc; since the binary is statically linked, there is no PLT/GOT.

Full breakdown with real-world context →

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-Oriented Programming (ROP) to call mprotect and then execute shellcode on the stack

Option C is correct because the binary is statically linked (no libc to return to) and has a non-executable stack (NX), so shellcode cannot execute directly on the stack. Return-Oriented Programming (ROP) allows the attacker to chain gadgets from the binary itself to call mprotect() and change the stack region to executable, then pivot to shellcode placed on the stack. Since ASLR is enabled but the binary is not PIE, its code base address is fixed, making ROP gadgets reliably addressable.

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.

✗
Heap spraying to place shellcode in the heap and then overwrite a function pointer to execute the shellcode
Why it's wrong here
Heap spraying is often used when ASLR is present, but NX still prevents execution from the heap unless the heap is made executable. Also, the binary is not PIE, so addresses are known, but the heap may still be non-executable.
✗
Return-to-libc attack using libc functions
Why it's wrong here
Return-to-libc is a technique against NX, but since the binary is statically linked (no libc loaded dynamically), the attacker would need to use gadgets from the binary itself, not libc.
✓
Return-Oriented Programming (ROP) to call mprotect and then execute shellcode on the stack
Why this is correct
ROP allows the attacker to chain gadgets to call mprotect and change memory permissions on the stack to executable, then jump to shellcode placed on the stack. This bypasses NX while leveraging the known addresses from the statically linked, non-PIE binary.
Clue confirmation
The clue word "most likely" in the question point toward this answer.
Related concept
Read the scenario before looking for a memorised answer.
✗
Ret2plt to call system() via the PLT
Why it's wrong here
Ret2plt is used against dynamically linked binaries to call functions from libc; since the binary is statically linked, there is no PLT/GOT.

Common exam traps

Common exam trap: answer the scenario, not the keyword

The trap here is that candidates assume return-to-libc is always viable, forgetting that a statically linked binary has no libc to return to, making ROP the only way to call mprotect and bypass NX.

Detailed technical explanation

How to think about this question

In a statically linked, non-PIE binary, all code (including potential ROP gadgets) resides at fixed addresses, which defeats ASLR for the binary's own segments. The mprotect() syscall (often via a gadget like 'pop rdi; ret' followed by 'pop rsi; ret' and 'pop rdx; ret') can set the stack page to RWX, allowing shellcode execution. A real-world example is exploiting a CGI binary in an embedded device where no libc is available and NX is enforced.

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.

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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-Oriented Programming (ROP) to call mprotect and then execute shellcode on the stack — Option C is correct because the binary is statically linked (no libc to return to) and has a non-executable stack (NX), so shellcode cannot execute directly on the stack. Return-Oriented Programming (ROP) allows the attacker to chain gadgets from the binary itself to call mprotect() and change the stack region to executable, then pivot to shellcode placed on the stack. Since ASLR is enabled but the binary is not PIE, its code base address is fixed, making ROP gadgets reliably addressable.

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.

Are there clue words in this question I should notice?

Yes — watch for: "most likely". Probability qualifier — the question wants the most probable cause or outcome, not a guaranteed one. Eliminate low-probability options.

What is the key concept behind this question?

Read the scenario before looking for a memorised answer.

About these practice questions

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Same concept, more angles

1 more ways this is tested on PT0-002

These questions test the same concept from different angles. Work through them to make sure you can recognise it however the exam phrases it.

Variation 1. A penetration tester gains a foothold on a Linux system with ASLR and NX enabled. The tester identifies a stack buffer overflow in a SUID binary. The binary has no PIE (Position Independent Executable) and is compiled without stack canaries. The tester wants to execute a shell. Which technique should be used?

hard

A.Return-to-libc attack
B.Heap spraying
✓ C.ROP chain
D.Buffer overflow with NOP sled

Why C: Since the binary has no PIE and lacks stack canaries, the attacker can predict the address of the return address on the stack. However, with ASLR and NX enabled, the stack is non-executable and system library addresses are randomized. A ROP chain allows the tester to bypass both protections by chaining small instruction sequences (gadgets) already present in the binary or loaded libraries to achieve arbitrary code execution, such as calling execve to spawn a shell.

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Question Discussion

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