# Is_int()

#1

Oops, try again. Your function fails on isint(-2). It returns False when it should return True._

I feel like I'm missing something that was referenced in a previous exercise, but I cannot tell what it is. I know that I have seen this code used before in a previous exercise, if anyone could point me towards it, I would be grateful.

``````def is_int(x):
if x == int():
return True
else:
return False``````

#2

Problem.

`if x == int():`

Can you explain what are you trying to do at this line of code?

#3

I was trying to get "if x is an integer, the following should return True." I think there is a better way of displaying this, however, and I know it was referenced before.

#4

If I had to check if x is an integer

I'd do something like this...

x = 4

`type(x) == int`

#5

Tried this code:

def is_int(x):
if type(x) == int:
return True
else:
return False

Error occurred:

Oops, try again. Your function fails on isint(7.0). It returns False when it should return True._

#6

Yep, You have to also check numbers like : 2.0, 45.00 , 67.00 (decimal part = 0),

because when you try that..

`type(34.0) # it will return float not int`

hint--

You can find decimal part of number and check if it is 0 or not!
If it is, then return True!
Can you think of any solution ?