rounding and floats

[Than999]

Hi, I don't understand why 10/3 the answer is 3 and 10.0/3.0 is 3.3333? Can anyone explain this to me? Thanks!

[bowlofred]

There's a python tutorial on the subject of floating point and rounding.

If you get 10/3 as 3, then that suggests you're using python 2, which has different semantics for floating point division and integer division. I suggest you start using python3 and learn the rules for that version instead.

In python2, the division result depends on the arguments (integer vs floating point). In python3, regular division can always be floating point if necessary. You use a different operator (//) to force integer results.

# python3
>>> 10/3
3.3333333333333335
>>> 10.0/3.0
3.3333333333333335

[deanhystad]

I wouldn't understand why 10/3 == 3 either. I cannot find a way to make 10/3 == 3. But I cannot do that because I am running Python 3.8. If I was running Python 2.7 10/3 does == 3.

In Python pre 3, dividing an integer by an integer produced an integer. That is why 10 / 3 == 3. It cannot equal 3.33333 because 3.3333 is not an integer. 3 the integer part left over when you remove the fractional .333333 part, so the answer is 3.

To do the same trick in Python 3 you need to add extra code. int(10/3) == 3.

By the way, integer division in C or C++ works the same way.

Things change when you start using floats. A float divided by anything produces a float and dividing anything by a float produces a float. That is why 10.0 / 3.0 == 3.333333 and 10.0 / 3 == 3.3333 and 10 / 3.0 == 3.33333. In this Python 2 and Python 3 agree. When you do math operations, if any of the operands is a float the result is a float.

By the way, floating point division in C or C++ works the same way.

There is a misconception that "//" is integer division in Python 3. This is incorrect. "//" is "floor division". Floor division is similar to integer division, but the result is not necessarily an integer.
10 // 3 == 3, 3 is an integer
10.0 // 3.0 == 3.0, 3.0 is a float if dividend or divisor are floats.