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Daniel Glennie
42,854 PointsSquaring function won't return correct answer
I have tried various minor adjustments to this code, using ValueError and TypeError in the except block and defining a ret variable to pass the return values to. I've tried using the isinstance method to check if num is of type int and also the type(int) method but I don't seem to be getting anywhere. I know it'll be something simple that I'm missing. Any ideas?
# EXAMPLES
# squared(5) would return 25
# squared("2") would return 4
# squared("tim") would return "timtimtim"
def squared(num):
try:
if type(num) is int:
return num * num
except TypeError:
return num * len(num)
2 Answers
Russell Sawyer
Front End Web Development Techdegree Student 15,705 PointsYou are pretty close. You just need to make a couple changes.
def squared(arg):
try:
arg = int(arg) #< this line determines if the variable is a number or integer
except ValueError:
return arg * len(arg)
else:
return arg ** 2
Patrick Liu
Full Stack JavaScript Techdegree Student 1,655 PointsWhy doesn't this work?
def squared(num):
try:
return int(num) * int(num) #<-- Doesn't this check to see if "num" is a integer too?
except ValueError:
print(len(num) * num)
Patrick Liu
Full Stack JavaScript Techdegree Student 1,655 Pointsdef squared(num):
try:
return int(num) * int(num)
except ValueError:
return len(num) * num
This works... it's cause I was printing and not returning the non integer value