PCAP Strings Practice Question
Exhibit
filepath = "C:\Users\Admin\Documents\file.txt"
with open(filepath, 'r') as f:
content = f.read()What is the likely outcome of running the following code?
```python
with open("C:\Users\path\file.txt", "r") as f:
print(f.read())```
⚠ Common exam trap
Python Institute often tests the misconception that backslashes in strings are always treated literally or that invalid escape sequences are silently ignored, leading candidates to choose options about file operations instead of recognizing the compile-time SyntaxError.
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
✓
The program raises a SyntaxError due to invalid escape sequences.
The code likely contains backslash sequences (e.g., `\U`, `\p`, or other non-standard escapes) that are not valid escape sequences in Python. Starting with Python 3.12, such invalid sequences raise a `SyntaxError` at compile time, preventing the program from running. In earlier versions, a `DeprecationWarning` is issued, but the code may run. The error is not about file operations or path resolution; it's a compile-time syntax error. To avoid this, use raw strings (`r"..."`) or double backslashes (`\\`).
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
The program raises a SyntaxError due to invalid escape sequences.
Why this is correct
A normal string literal is parsed by the Python compiler before any code executes, and every backslash must form a legal escape sequence. `\U` is the escape prefix for a 32-bit Unicode code point and must be followed by exactly eight hexadecimal digits; here it is truncated, so the tokenizer raises `SyntaxError: (unicode error) ... truncated \UXXXXXXXX escape` at compile time. Consequently, no program output or file operation ever occurs.
- ✗
The program runs but the file contents are incorrect.
Why it's wrong here
This option incorrectly assumes execution reaches runtime. Because the invalid `\U` escape is detected while the source is being tokenized and compiled, Python never builds bytecode and never invokes the `open()` call, so no writes can take place and no file contents can be created or altered. The failure mode is a compile-time `SyntaxError`, not a successful run with bad data.
- ✗
The program raises a FileNotFoundError because the path is invalid after escape interpretation.
Why it's wrong here
`FileNotFoundError` is a runtime OSError raised by the operating system when `open()` cannot locate a path, but that call is never reached. The path argument is irrelevant here: the `\U` sequence is rejected during parsing, before Python evaluates the expression or calls any built-in function. Even if a valid escape sequence had changed the path, the exception would still occur later and would not be the first failure.
- ✗
The file is opened successfully because backslashes are ignored.
Why it's wrong here
Backslashes inside a non-raw Python string literal are active escape characters; Python does not silently ignore them. A single backslash before `U` initiates the `\U` escape, and an incomplete `\U` sequence is a compile-time error, so the file object is never created. To treat backslashes literally, the source would need either a raw string such as `r'...'` or doubled backslashes (`\\`).
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