Is_int problem!


#1



I've already tried this codes but i keep getting this error:
Oops, try again. Your function fails on is_int(-2). It returns None when it should return True.
the codes:
1.
def is_int(x):
if round(x)==x:
print True
else:
print False
2.
def is_int(x):
if type(x)==int:
print True
else:
print False
3.
def is_int(x):
if round(x)-x==0:
print True
else:
print False


Replace this line with your code.


#2

The operative word here is return.


#4

How did you know that you should use round() ???


#5

Ok here instead of using the round() function in your code try my code which is given below and check whether you get a go ahead as i got and also if u think mark it as a solution :smile:

def is_int(x):
temp = x - int(x)
if temp == 0:
return True
else:
return False

for the explanation:
here we first calculate the value of temp by subtracting the integer part of x with original value of x (explicit type conversion as in code). Then if temp >0 it means there is a fractional part and hence the number is not an integer and return False OTHERWISE return True


Is_int
#6

thanks for your help buddy


#7

Thank U:slight_smile:


#8

Most Welcome Bro :smile:


#9

You answer is really clever good job :wink:


#10

There is only one problem. The instructions specifically state that we are not to use the int() function in our solution. If clever means disobey the instructions, then I guess it can be called a good job. Recommend find another solution that does not use, int().


#11