PCAP 31 03 PCAP Certified Associate Python Programmer · Free Practice Question Medium

Question 25

Q213 - Functions


What is the expected output of the following code?



  • A
  • B
  • C
  • D
  • E
Reveal correct answer

Correct answer: E

Explanation

Topics: def default parameter multiply operator string concatenation

Try it yourself:

Explanation:

In the function a string concatenation by multiplication takes place.

Once with the default value of num (1)

and once with the one passed by argument (3)


Q213 (Please refer to this number, if you want to write me about this question.)

A. This choice is incorrect because it includes a comma and the word "Viewers" after 'Hello', which is not part of the expected output. The function calls only print the messages 'Hello' and 'WelcomeWelcomeWelcome'.

B. This choice is incorrect because it includes commas between the words 'Welcome', which is not part of the expected output. The function calls only print the messages 'Hello' and 'WelcomeWelcomeWelcome' without any commas.

C. This choice is incorrect because it includes spaces between the words 'Welcome', which is not part of the expected output. The function calls only print the messages 'Hello' and 'WelcomeWelcomeWelcome' without any spaces between the repeated 'Welcome' words.

D. This choice is incorrect because it only includes the output for the first function call func('Hello'), which prints 'Hello' once. It does not include the output for the second function call func('Welcome', 3), which prints 'Welcome' three times.

E. The function func takes two parameters, message and num, with a default value of 1 for num. When the function is called with func('Hello'), it prints the message 'Hello' once. When the function is called with func('Welcome', 3), it prints the message 'Welcome' three times. Therefore, the expected output is 'Hello' followed by 'WelcomeWelcomeWelcome'.

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