round behaviour differs between python2 and python3

Question

HELP Im currently using a book, but it seems like its wrong or something. This is what I enter:

round(10)
10
>>> round(10.0)
10
>>> round(10.2)
10
>>> round(8.7)
9
>>> round(4.5, 1)
4.5
>>> round(4.5, 2)
4.5
>>> round(4.5, 3)
4.5
>>> round (4.5)
4
>>> round(4.5)
4

Here is what the book says should happen:

round(10)
10
>>> round(10.0)
10.0
>>> round(10.2)
10.0
>>> round(8.7)
9.0
>>> round (4.5)
5

• juanpa.arrivillaga: You are probably using Python 3, and the book is for Python 2.
• aquil.abdullah: Forget the book. Look at what round is supposed to do...You can get the behavior described by the book by supplying the number of significant digits. round(10.0, 1) --> 10.0. The docs say "digits (default 0 digits)." NOTE: This is for Python 2.7x
• coldspeed95: @LOL, LOL......
• cdarke: Was the book for python 2 or python 3? Were you using python 2 or 3? My guess - the book was written for python 2 and you are using python 3.
• LOL: Actually, we are both using the same version of python...

Answer 1

Python2.7

>>> round(10.0)
10.0
>>> round(10.0, 0)
10.0

Python3.4, 3.5, and 3.6

>>> round(10.0)
10
>>> round(10.0, 0)
10.0
>>> 

Two takeaways:

1. round in python3 returns an int (when it can) while in python2 returns float (always).
2. Read a book for python3!

• anon: For me round(10.0) is returning 10. Using Python3.6.1
• LOL: Ok...im using 3.5.1, is that it?
• coldspeed95: @LOL Can confirm, 3.5 shows the same behaviour. Now you can safely assume you're reading the wrong book. Cheers.