function is not returning what is intended

Question

The function that I created is to measure the length of an argument!. I want to return "Sorry, integers don't have length"

def string_length(mystring):
 if type(mystring) == int:
 return "Sorry, integers don't have length"
 else:
 return len(mystring)
mystring = input("what is the value ?")
print(string_length(mystring))

what am I doing wrong?

• user2357112: You didn't pass it an int.

Answer 1

Your question isn't very clear, but this should work, if my understanding is not wrong:

def string_length(mystring):
 try:
 if type(int(mystring)) == int:
 return "Sorry, integers don't have length"
 elif type(mystring) == str:
 return len(mystring)
 except:
 return len(str(mystring))
mystring = input("what is the value ?")
print(string_length(mystring))

• OneCricketeer: type(int(mystring)) == int is always true, though. I think you just masked the problem with a try-except

Answer 2

In Python2, input will evaluate to an integer, not a string, which would explain the message.

You can use hasattr to check for the __len__ magic function rather than doing explict type checking.

This should work for any iterable type, I believe

def string_length(mystring):
 if hasattr(mystring, '__len__'):
 return len(mystring)
 else:
 print("Sorry, {} don't have length".format(type(mystring)))
 return None
mystring = input("what is the value ?")
print(string_length(mystring))

Then, make sure you are running this in Python3