Python Argument Parsing MCQ

This set of Python Multiple Choice Questions & Answers (MCQs) focuses on “Argument Parsing”.

1. What is the type of each element in sys.argv?
a) set
b) list
c) tuple
d) string
View Answer

Answer: d
Explanation: sys.argv is a list in Python that contains command-line arguments passed to the script. Each element in sys.argv is of type string, regardless of whether it looks like a number or not. The first element (sys.argv[0]) is the script name.

2. What is the length of sys.argv?
a) number of arguments
b) number of arguments + 1
c) number of arguments – 1
d) none of the mentioned
View Answer

Answer: b
Explanation: sys.argv includes the name of the script as the first element (sys.argv[0]), followed by the actual command-line arguments. So, if there are n command-line arguments, the length of sys.argv will be number of arguments + 1.

3. What will be the output of the following Python code?

def foo(k):
    k[0] = 1
q = [0]
foo(q)
print(q)

a) [0]
b) [1]
c) [1, 0]
d) [0, 1]
View Answer

Answer: b
Explanation: Lists in Python are mutable, so when the list q is passed to the function foo, the change made (k[0] = 1) modifies the original list. Therefore, after the function call, q becomes [1].

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4. How are keyword arguments specified in the function heading?
a) one-star followed by a valid identifier
b) one underscore followed by a valid identifier
c) two stars followed by a valid identifier
d) two underscores followed by a valid identifier
View Answer

Answer: c
Explanation: In Python, keyword arguments (also known as **kwargs) are specified in the function heading using two asterisks (**) followed by a valid identifier. For example:

def func(**kwargs):
    print(kwargs)

This allows the function to accept an arbitrary number of keyword arguments as a dictionary.

5. How many keyword arguments can be passed to a function in a single function call?
a) zero
b) one
c) zero or more
d) one or more
View Answer

Answer: c
Explanation: In Python, a function can accept zero or more keyword arguments during a single call. These keyword arguments are specified by explicitly naming the parameters (e.g., func(a=1, b=2)), and Python functions can handle any number of such arguments, including none.

6. What will be the output of the following Python code?

def foo(fname, val):
    print(fname(val))
foo(max, [1, 2, 3])
foo(min, [1, 2, 3])

3
1

1
3

c) error
d) none of the mentioned
View Answer

Answer: a
Explanation: The function foo takes a function fname and a value val as arguments. It then applies fname to val and prints the result.

foo(max, [1, 2, 3]) applies max() to the list, which returns 3.
foo(min, [1, 2, 3]) applies min() to the list, which returns 1.

So, the output is:

3
1

7. What will be the output of the following Python code?

def foo():
    return total + 1
total = 0
print(foo())

a) 0
b) 1
c) error
d) none of the mentioned
View Answer

Answer: b
Explanation: The function foo() returns total + 1. Even though total is not defined inside the function, it is defined in the global scope before the function is called. Python allows reading a global variable inside a function unless you try to assign to it. So total = 0 globally, and foo() returns 0 + 1 = 1.

8. What will be the output of the following Python code?

def foo():
    total += 1
    return total
total = 0
print(foo())

a) 0
b) 1
c) error
d) none of the mentioned
View Answer

Answer: c
Explanation: The code results in an error because inside the function foo(), you are trying to modify the variable total without declaring it as global. Python treats total as a local variable, but it’s being used before assignment, leading to an UnboundLocalError.

9. What will be the output of the following Python code?

def foo(x):
    x = ['def', 'abc']
    return id(x)
q = ['abc', 'def']
print(id(q) == foo(q))

a) True
b) False
c) None
d) Error
View Answer

Answer: b
Explanation: The function foo(x) assigns a new list [‘def’, ‘abc’] to x, so x now points to a different object in memory. Therefore, the id(x) inside foo() will differ from id(q) outside the function, and id(q) == foo(q) will return False.

10. What will be the output of the following Python code?

def foo(i, x=[]):
    x.append(i)
    return x
for i in range(3):
    print(foo(i))

[0]
[1]
[2]

[0]
[0, 1]
[0, 1, 2]

[1]
[2]
[3]

d) None of the mentioned
View Answer

Answer: b
Explanation: The default argument x=[] is evaluated only once when the function is defined, not each time it’s called. So the same list is used and modified across all calls to foo(). As a result, each call appends to the same list, and the list grows with each iteration.

[0]
[0, 1]
[0, 1, 2]

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