What is the expected output of the following code? def func(message, num=1): print(message * num) func('Hello') func('Welcome', 3)

The correct answer is D. 1 | Hello 2 | Welcome Welcome Welcome. Option D is correct because Python's string multiplication operator (`) repeats the string without any separator - 'Welcome' 3 produces 'WelcomeWelcomeWelcome', and 'Hello' 1 (the default) produces 'Hello'. Why distractors fail: A is wrong on the count - it shows Welcome only…

Question

What is the expected output of the following code?
def func(message, num=1):
print(message * num)
func('Hello')
func('Welcome', 3)

Options

A1 | Hello 2 | Welcome Welcome
B1 | Hello 2 | Welcome Welcome Welcome
C1 | Hello 2 | Welcome, Welcome, Welcome
D1 | Hello 2 | Welcome Welcome Welcome
E1 | Hello

How the community answered

(46 responses)

B
11% (5)
C
4% (2)
D
83% (38)
E
2% (1)

Explanation

Option D is correct because Python's string multiplication operator (*) repeats the string without any separator - 'Welcome' * 3 produces 'WelcomeWelcomeWelcome', and 'Hello' * 1 (the default) produces 'Hello'.

Why distractors fail:

A is wrong on the count - it shows Welcome only twice, not three times.
B looks close but inserts spaces between repetitions (Welcome Welcome Welcome), which * never does.
C incorrectly adds commas, as if using ', '.join(...) - that's a completely different operation.
E omits the second line of output entirely, ignoring func('Welcome', 3).

Memory tip: Think of string multiplication as "paste n copies side by side with no glue" - 'ab' * 3 → 'ababab', never 'ab ab ab'. If you want separators, you need join(), not *.

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