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

You are probably using Python 3, and the book is for Python 2. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Jun 21, 2017 at 19:07
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 — aquil.abdullah
– aquil.abdullah, Commented Jun 21, 2017 at 19:07
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. — cdarke
– cdarke, Commented Jun 21, 2017 at 19:08

cs95 · Accepted Answer · 2017-06-21 19:17:34Z

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:

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

edited Jun 21, 2017 at 19:17

answered Jun 21, 2017 at 19:09

cs95

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3 Comments

anon Over a year ago

For me round(10.0) is returning 10. Using Python3.6.1

LOL Over a year ago

Ok...im using 3.5.1, is that it?

cs95 Over a year ago

@LOL Can confirm, 3.5 shows the same behaviour. Now you can safely assume you're reading the wrong book. Cheers.

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