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Start your free trialArthur Berg
11,350 Pointstry, except, else
I can't see what's wrong with my code? I'm checking if the argument is an int with type(num) == int. Is this the wrong way to do it?
Appreciate any help!
# EXAMPLES
# squared(5) would return 25
# squared("2") would return 4
# squared("tim") would return "timtimtim"
def squared(num):
try:
type(num) == int
except ValueError:
return num * len(num)
else:
return num * num
4 Answers
tobiaskrause
9,160 Pointsdef squared(num):
try:
int(num)
except ValueError:
return num * len(num)
else:
return int(num) ** 2
// or return int(num) * int(num)
Arthur Berg
11,350 Pointsahh I forgot to convert it , obvs... my type(num) == int syntax works like int(num) in the try block?
tobiaskrause
9,160 PointsWell let me show you on IDLE:
>>> int("1")
1
>>> int("test")
Traceback (most recent call last):
File "<pyshell#1>", line 1, in <module>
int("test")
ValueError: invalid literal for int() with base 10: 'test'
>>>
>>> num = "test"
>>> type(num) == int
False
>>>
You get the difference ;)?
type() returns True or False while int() returns an integer (string converted to an int) or (what we need) an error-message
Arthur Berg
11,350 PointsIt didn't !
tobiaskrause
9,160 Pointsdont copy my comment Just to show you some other ways
// or return int(num) * int(num)
Arthur Berg
11,350 PointsI get it, thanks Tobias! New to Python, learning it in order to build shopping system with stripe and django. Coming from beginner/intermediate level of Javascript, mostly angular!