Question

What type of error is shown and what is the fix?

1. Print("Hello world! ")
2. fav_num = 3
3. if fav_num == 1:
4. print("1 is your favorite number")
5. if fav_num == 2:
6. print("2 is your favorite number")
7. if fav_num == 3:
8. print("3 is your favorite number")
9. if fav_num == 4:
10. print("4 is your favorite number")
11. else:
12. print("Your favorite number is not 1, 2, 3, or 4")

Hello world!
3 is your favorite number
Your favorite number is not 1, 2, 3, or 4

(1 point)
Responses

Runtime Error; Fix the syntax on line 3, so that we do not encounter runtime issues.

Syntax Error; When comparing variable and their expected values, you need to use the comparison operator '==' instead of '='

Logic Error; The 'else' statement on line 11 will fire if the 'if' statement on line 9 only is false. To get a single result from the 'if' and 'else' statements in lines 3 through 12, you must change 'if' to 'elif' on lines 5, 7, and 9.

Syntax Error; You cannot use '_' in a variable name when nesting into an 'if' statement.

Answer 1

This is a logic error. Because each test is a separate if, the final else only attaches to the last if and therefore runs even when an earlier if was true. Change the later ifs to elif (or use an if/elif/else chain) so only one branch runs.

Fixed code:

print("Hello world!")
fav_num = 3
if fav_num == 1:
print("1 is your favorite number")
elif fav_num == 2:
print("2 is your favorite number")
elif fav_num == 3:
print("3 is your favorite number")
elif fav_num == 4:
print("4 is your favorite number")
else:
print("Your favorite number is not 1, 2, 3, or 4")

Correct output:
Hello world!
3 is your favorite number