PCEP-30-02: Block 4: Functions and Exceptions

Topic snapshot

Sample questions

These questions are original IT Mastery practice items aligned to this topic area. They are designed for self-assessment and are not official exam questions.

Question 1

Topic: Block 4: Functions and Exceptions

Examine this code:

price = 50

def show_price(price):
    print(price)

show_price(30)
print(price)

Which statement is correct?

Options:

• A. The price parameter shadows the outer name; output is 30 then 50.

• B. Only the outer price is used; output is 50 then 50.

• C. Using price globally and as a parameter causes an error.

• D. The price parameter updates the outer name; output is 30 then 30.

Best answer: A

Explanation: Function parameters are local to the function. Here, the parameter price receives 30 and hides the outer price only inside show_price(), so the function prints 30 while the global variable stays 50.

Shadowing happens when a local name inside a function matches a name from an outer scope. In this example, show_price(price) creates a local parameter named price. When the function is called with 30, that local parameter gets the value 30, and it hides the global price only while the function runs. It does not replace or modify the global variable, because the parameter is a separate local name. After the function finishes, print(price) refers again to the global variable, which is still 50. The key idea is that the same spelling does not mean the same variable in every scope.

• The option claiming the outer name becomes 30 confuses passing an argument with changing a global variable.
• The option claiming both outputs are 50 ignores that the function uses its local parameter first.
• The option claiming an error is wrong because Python allows a parameter to have the same name as a global variable.

Question 2

Topic: Block 4: Functions and Exceptions

What is printed by this Python 3 code?

try:
    print(int("7a"))
except Exception:
    print("general")
except ValueError:
    print("value")

Options:

• A. The program stops with ValueError

• B. general

• C. value

• D. generalthenvalue

Best answer: B

Explanation: Python checks except clauses from top to bottom and runs the first matching handler. int("7a") raises ValueError, but ValueError is a subclass of Exception, so the broader handler runs first and prints general.

In a try/except statement, handler order matters. Python does not search for the most specific match first; it uses the first except clause that can handle the raised exception.

• int("7a") raises ValueError.
• ValueError is a subclass of Exception.
• The except Exception clause appears first, so it handles the error.
• The later except ValueError clause is never reached.

That is why the only output is general. To avoid hiding specific handlers, place more specific exceptions before broader ones.

• The option showing value assumes Python automatically prefers the most specific handler, but it does not.
• The option showing both outputs fails because only one matching except block runs for one exception.
• The option claiming the program stops with ValueError ignores that the earlier except Exception already catches it.

Question 3

Topic: Block 4: Functions and Exceptions

A student writes this function:

def report(title, pages=1, color=False):
    pass

Which TWO calls are invalid because the same parameter gets a value more than once?

Options:

• A. report("Guide", title="Notes")

• B. report("Guide", 3, color=True)

• C. report(title="Guide", pages=3)

• D. report(color=True, title="Guide")

• E. report("Guide", pages=3, color=True)

• F. report("Guide", 3, pages=4)

Correct answers: A and F

Explanation: A parameter can receive a value by position or by keyword, but not both in the same call. The invalid calls are the ones where pages or title is assigned positionally first and then assigned again with a keyword.

Python matches positional arguments first, from left to right, and then applies keyword arguments. A call is invalid when one parameter ends up receiving two values.

In report("Guide", 3, pages=4), the value 3 already fills pages as the second positional argument, so pages=4 tries to fill pages again. In report("Guide", title="Notes"), the first positional argument already fills title, so title="Notes" duplicates that parameter.

Calls that use only keywords, only positionals, or a mix that targets different parameters are valid. The key check is simple: after matching the arguments, no parameter may have been assigned twice.

• report("Guide", pages=3, color=True) is valid because title, pages, and color are each supplied once.
• report(title="Guide", pages=3) is valid because both arguments are keywords and neither parameter is repeated.
• report(color=True, title="Guide") is valid because keyword arguments can appear in any order when each parameter is used once.
• report("Guide", 3, color=True) is valid because title and pages are filled positionally, and color is filled once by keyword.

Question 4

Topic: Block 4: Functions and Exceptions

A beginner runs this program. Assume there is no other try/except in the program.

def pick(items, n):
    return items[n]

def show(items, n):
    try:
        print(pick(items, n))
    except KeyError:
        print("missing key")

names = ["Ann", "Bob"]
show(names, 2)

The program crashes. Which statement best explains the crash?

Options:

• A. pick() returns None, and print() crashes on that value.

• B. pick() raises IndexError, but show() only catches KeyError.

• C. The except KeyError block catches the error and prints missing key.

• D. pick() raises KeyError because list position 2 is missing.

Best answer: B

Explanation: names[2] is outside the valid list indexes, so Python raises IndexError. The active handler in show() is only for KeyError, so the exception is not caught and continues upward until the program stops.

Python handles an exception only when an active except clause matches the actual exception type. Here, pick(items, n) tries to read names[2], but the list has only two elements, so the real error is IndexError. The try block in show() is active, but its handler is except KeyError, which is for missing dictionary keys, not out-of-range list indexes.

Because no matching handler exists in show(), the IndexError propagates back to the caller. The stem says there is no other try/except, so the exception remains unhandled and the program crashes. The key idea is to identify both the real exception type and whether any active handler matches it.

• The option claiming a missing list position raises KeyError confuses list indexing with dictionary lookup.
• The option saying missing key is printed ignores that except KeyError does not match IndexError.
• The option saying pick() returns None is wrong because out-of-range indexing raises an exception before any value is returned.

Question 5

Topic: Block 4: Functions and Exceptions

A student is making study notes and stores exception names as strings in a dictionary.

notes = {
    "root": "?",
    "examples": ["TypeError", "ValueError", "ZeroDivisionError"]
}

Which string should replace ? to make the note about Python’s built-in exception hierarchy correct?

Options:

• A. RuntimeError

• B. ArithmeticError

• C. Exception

• D. BaseException

Best answer: D

Explanation: In Python’s built-in hierarchy, BaseException sits above Exception and all standard exception branches. Since the dictionary stores class names as strings, the root entry must be BaseException.

BaseException is the top-level built-in exception base class in Python. Most exceptions beginners commonly use, such as TypeError, ValueError, and ZeroDivisionError, are descendants of Exception, and Exception itself inherits from BaseException. That makes BaseException the only correct value for the dictionary’s root key.

A simple way to view the hierarchy is:

• BaseException
• Exception
• specific subclasses such as TypeError and ValueError

The closest distractor is Exception, because it is a very common base class in try/except code, but it is still one level below BaseException.

• Exception is close but lower: it is a major base class, but it inherits from BaseException.
• ArithmeticError is a branch: it groups numeric-related exceptions and is not the root of all built-in exceptions.
• RuntimeError is too specific: it is one ordinary exception class, not a top-level base class.

Question 6

Topic: Block 4: Functions and Exceptions

Consider this function:

def fetch(data, item):
    return data[item]

If numbers is a list, fetch(numbers, 10) can raise IndexError. If prices is a dictionary, fetch(prices, "milk") can raise KeyError when the key is missing. Which except clause catches both through their common built-in exception family?

Options:

• A. except ArithmeticError:

• B. except TypeError:

• C. except ValueError:

• D. except LookupError:

Best answer: D

Explanation: Use except LookupError: because both examples are lookup failures caused by data[item]. Python places IndexError and KeyError under the same built-in exception family, so one handler can catch either case.

When a function uses data[item], Python performs a lookup in a container. If the container is a list and the index does not exist, Python raises IndexError. If the container is a dictionary and the key is absent, Python raises KeyError. Those two specific exceptions share the same parent class: LookupError.

That is why except LookupError: can handle both situations with one clause. The other choices name different exception families: ArithmeticError is for math-related errors, ValueError is for an unacceptable value, and TypeError is for using an operation with an incompatible type. The key takeaway is that failed indexing and failed key access are both lookup-related exceptions.

• Arithmetic family is unrelated because the function is not doing a numeric operation.
• Bad value is not the right match here; missing keys and invalid indexes are classified as lookup failures.
• Wrong type would apply to an incompatible type being used, not to an absent element in a valid container.

Question 7

Topic: Block 4: Functions and Exceptions

What is printed by this Python 3 code?

def total(n):
    if n <= 0:
        return 0
    return n + total(n - 2)

print(total(5))

Options:

• A. 9

• B. 15

• C. 8

• D. RecursionError

Best answer: A

Explanation: This recursive function has a clear base case: when n <= 0, it returns 0. Starting from 5, each call adds the current value and then calls itself with n - 2, producing 5 + 3 + 1 + 0.

In recursion, you need a base case and a recursive case. Here, the base case is n <= 0, which stops the function and returns 0. The recursive case returns the current n plus the result of calling the function again with n - 2.

So the calls unfold like this:

• total(5) returns 5 + total(3)
• total(3) returns 3 + total(1)
• total(1) returns 1 + total(-1)
• total(-1) returns 0

When the calls finish, the values are added during the return phase: 5 + 3 + 1 + 0 = 9. The closest wrong idea is adding every number from 1 to 5, but this function skips even numbers because it subtracts 2 each time.

• One term missed The value 8 ignores one of the recursive additions, but both 3 and 1 are included before the base case returns 0.
• Adds every number The value 15 would come from 5 + 4 + 3 + 2 + 1, but the function decreases by 2, not by 1.
• No stop condition RecursionError would happen without a reachable base case, but n <= 0 is reached after a few calls.

Question 8

Topic: Block 4: Functions and Exceptions

A beginner wrote this program:

values = ["3", "two", "4"]
total = 0

for v in values:
    total += int(v)

print(total)
print("done")

The goal is to skip any value that cannot be converted with int() and continue processing the rest of the list. Which revision is the best fix?

Options:

• A. Replace the loop with:

for v in values:
    try:
        total += int(v)
    except ValueError:
        continue

• B. Replace the loop with:

for v in values:
    if v:
        total += int(v)

• C. Replace the loop with:

for v in values:
    try:
        total += int(v)
    except TypeError:
        continue

• D. Replace the loop with:

try:
    for v in values:
        total += int(v)
except ValueError:
    pass

Best answer: A

Explanation: int("two") raises a ValueError, so the risky conversion must be handled with try-except. Putting that handler inside the loop lets the program skip only the bad item and keep processing later values.

The core idea is to catch the specific runtime exception at the point where it can happen. Here, int(v) is the risky operation, and a non-numeric string like "two" raises ValueError.

When the try-except is inside the for loop, each list element is handled independently. If one conversion fails, the except ValueError block runs and continue moves to the next iteration, so later values such as "4" are still added to total.

If the whole loop is wrapped in one try, the first bad value ends the loop early. Catching TypeError does not help because the actual exception is ValueError. Checking whether a string is truthy only tests whether it is empty, not whether it contains digits.

Handle the smallest risky statement with the correct exception type when you want the loop to keep running.

• Whole-loop handler catches the error after the loop has already stopped, so later values are not processed.
• Wrong exception type fails because int("two") raises ValueError, not TypeError.
• Truthiness check only rejects empty strings; a non-empty invalid string still crashes at int(v).

Question 9

Topic: Block 4: Functions and Exceptions

What is printed when this Python code runs?

def third():
    print("third")
    return 10 / 0

def second():
    print("second")
    third()
    print("after")

def first():
    try:
        second()
    except ZeroDivisionError:
        print("handled")

first()

Options:

• A. second, third, after, handled

• B. second, handled

• C. second, third, then an unhandled ZeroDivisionError

• D. second, third, handled

Best answer: D

Explanation: The error happens inside third(), but neither third() nor second() handles it. The exception propagates back to first(), where the except ZeroDivisionError block runs, so after is never printed.

An unhandled exception moves outward through function calls until Python finds a matching except block or the program ends. Here, second() calls third(), and third() prints third before raising ZeroDivisionError during 10 / 0.

• second() prints second
• third() prints third
• the exception leaves third() and then second()
• first() catches it and prints handled

Because the call to third() does not finish normally, control never reaches print("after") in second(). The key idea is that exceptions cross function boundaries when they are not handled locally.

• The option including after fails because execution in second() stops as soon as third() raises the exception.
• The option omitting third fails because that line is printed before the division error occurs.
• The option claiming the error is unhandled fails because first() has a matching except ZeroDivisionError block.

Question 10

Topic: Block 4: Functions and Exceptions

What is the output of this program?

def check():
    print("check")
    return False

if check:
    print("go")
else:
    print("stop")

Options:

• A. Only go is printed.

• B. Only stop is printed.

• C. check and then stop are printed.

• D. check and then go are printed.

Best answer: A

Explanation: The code uses check instead of check(), so Python treats it as a reference to the function object, not a function call. The function body never runs, and the condition is truthy, so only go is printed.

In Python, a function name by itself refers to the function object. To execute the function, you must add parentheses, such as check(). Here, the condition is if check:, so Python does not run the function body at all.

Because the function is never called:

• print("check") does not execute.
• return False is never reached.
• The condition tests the function object itself.
• A function object is truthy, so the if branch runs.

That is why the program prints only go. The closest mistake is reading if check: as though it were if check():.

• The option claiming check and then go fails because the function body never runs without parentheses.
• The option claiming check and then stop fails because return False is never reached.
• The option claiming only stop fails because a function object used in a condition is truthy.

Continue with full practice

Use the PCEP-30-02 Practice Test page for the full IT Mastery route, mixed-topic practice, timed mock exams, explanations, and web/mobile app access.

Try PCEP-30-02 on Web View PCEP-30-02 Practice Test

Related focused pages

• Free PCEP-30-02 Full-Length Practice Exam
• Python Fundamentals
• Control Flow
• Data Collections

Free review resource

Read the PCEP-30-02 Cheat Sheet on Tech Exam Lexicon, then return to IT Mastery for timed practice.