Welcome to the Treehouse Community
Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.
Looking to learn something new?
Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.
Start your free trialmamadou diene
Python Web Development Techdegree Student 1,971 PointsThis challenge is similar to an earlier one. Remember, though, I want you to practice! You'll probably want to use try a
Im stuck on this challenge
# EXAMPLES
# squared(5) would return 25
# squared("2") would return 4
# squared("tim") would return "timtimtim"
def square(argum):
try:
if argum = int:
return argum*argum
except ValueError:
return len(argum)* argum
3 Answers
Chris Freeman
Treehouse Moderator 68,454 PointsYou have the right idea. To check is something is an int
you can use isinstance(argum, int)
to test object type. (Taught in later lesson)
Since you have a try
block, you can also remove the if
and just "go for it" by using argum = int(argum)
Remember to align the except
with the try
Post back if you need more help. Good luck!!!
mamadou diene
Python Web Development Techdegree Student 1,971 Pointsthat is after i remove the "if", and correct the rest of the line like you said.
mamadou diene
Python Web Development Techdegree Student 1,971 PointsIt worked!!!!! Thank you!!!!
mamadou diene
Python Web Development Techdegree Student 1,971 Pointsmamadou diene
Python Web Development Techdegree Student 1,971 PointsThis is what i am getting now: SyntaxError: argum = int(argum):
Chris Freeman
Treehouse Moderator 68,454 PointsChris Freeman
Treehouse Moderator 68,454 PointsThe colon should be removed