A menu loop that controls user interaction

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Quick Questions

Shade in the A or B circle for each question.

1. Which loop header keeps a menu running until the user types q to quit?

A while choice != "q":

B while choice == "q":

2. What is the correct order of steps inside one pass of a menu loop?

A Read a choice, call the function, then show the menu

B Show the menu, read a choice, call that option's function

3. A user types "x", which is not a menu option. What should the menu loop do?

A Print an invalid choice message and show the menu again

B End the program straight away

4. Which line belongs in the loop to handle the quit option?

A elif choice == "q": print("Goodbye")

B elif choice == "q": choice = "a"

5. Why must the choice be read INSIDE the while loop?

A So the menu is only shown once

B So the user can pick a new option on every repeat

Match the Pairs

Draw a line connecting each item on the left to its match on the right.

Left column		Right column
1while choice != "q":		AKeeps the menu repeating until the user types q to quit
2else: print("Invalid choice")		BThere is no else branch, so an invalid choice is ignored silently. Add else: print("Invalid choice") so the user knows to try again.
3Each valid menu option calls its own function inside the loop		Cwhile choice != "q": choice = input("Choice: ")
4The menu must read a new choice on every pass so the user can quit		DShows an invalid choice message when the input matches no menu option, then the menu appears again
5while choice != "q": show_menu() if choice == "1": view_students()		Eif choice == "1": view_students()
6while choice != "q": choice = input("Choice: ") if choice == "1": view_students() elif choice == "2": add_student()		FThe choice is never read inside the loop. choice never changes, so the loop never ends and the user cannot quit. Add choice = input(...) after show_menu().

Definitions

while choice != "q":

Fill in the Blanks

Complete each line of code by filling in the missing parts.

choice = "" _____ choice != "q": print_menu()

choice = input("Choice: ") while choice __ "q":

____ choice == "2": show_students()

____: print("Invalid choice, try again")

choice = _____("Enter choice: ") run_option(choice)

Spot the Error

Shade in the letter of the line that contains an error.

Question 1

A choice = ""

B while choice == "q":

C choice = input("Option: ")

D if choice == "1":

E view_items()

Question 2

A choice = ""

B while choice != "q":

C show_menu

D choice = input("> ")

Question 3

A choice = ""

B while choice != "q":

C choice = int(input("> "))

D if choice == "1":

E add_item()

Question 4

A def main:

B choice = ""

C while choice != "q":

D choice = input("> ")

Question 5

A choice = ""

B while choice != "q":

C choice = input("> ")

D if choice == 1:

E view_items()

F elif choice == "2":

Answer Key

A menu loop that controls user interaction

Section A: Quick Questions

A — while choice != "q":

B — Show the menu, read a choice, call that option's function

A — Print an invalid choice message and show the menu again

A — elif choice == "q": print("Goodbye")

B — So the user can pick a new option on every repeat

Section B: Match the Pairs

while choice != "q": → Keeps the menu repeating until the user types q to quit

else: print("Invalid choice") → Shows an invalid choice message when the input matches no menu option, then the menu appears again

Each valid menu option calls its own function inside the loop → if choice == "1": view_students()

The menu must read a new choice on every pass so the user can quit → while choice != "q": choice = input("Choice: ")

while choice != "q": show_menu() if choice == "1": view_students() → The choice is never read inside the loop. choice never changes, so the loop never ends and the user cannot quit. Add choice = input(...) after show_menu().

while choice != "q": choice = input("Choice: ") if choice == "1": view_students() elif choice == "2": add_student() → There is no else branch, so an invalid choice is ignored silently. Add else: print("Invalid choice") so the user knows to try again.

Section C: Definitions

while choice != "q": — A menu loop repeats: show the menu, read a choice, call that option's function, and stop only when the user picks quit. Other input gets an invalid choice message.

Section D: Fill in the Blanks

choice = ""
while choice != "q":
    print_menu()

choice = input("Choice: ")
while choice != "q":

elif choice == "2":
    show_students()

else:
    print("Invalid choice, try again")

choice = input("Enter choice: ")
run_option(choice)

Section E: Spot the Error

B — The loop must keep running until the user picks quit, so the condition should be != "q". With == "q" the loop never starts because choice begins as an empty string.

C — A function is only called when it has brackets. Without them the menu is never shown. It should be show_menu().

C — int() would crash when the user types q to quit, and a number would never equal the string "q". Menu choices should stay as strings: choice = input("> ").

A — A function definition needs brackets after its name. It should be def main():.

D — input() returns text, so choice is the string "1". Comparing it to the number 1 is never true and the option's function would never run. It should be if choice == "1":.