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 trialPeter Falcone
Courses Plus Student 1,108 Pointssquared.py
I've tried numerous things. I feel the problem is likely the line just below 'try:' where I have changed out several different things.
# EXAMPLES
# squared(5) would return 25
# squared("2") would return 4
# squared("tim") would return "timtimtim"
def squared(num):
try:
int(num)
return(num * num)
except ValueError:
return(num * len(num))
2 Answers
Ryan S
27,276 PointsHi Peter,
You are really close, but the issue is that calling int(num)
on its own doesn't permanently convert the number to an integer. You need to assign it back to "num" so that the variable gets updated and can be successfully multiplied in your return statement.
def squared(num):
try:
num = int(num)
return(num * num)
except ValueError:
return(num * len(num))