EOL - 11 months ago 137

Python Question

The

`//`

`>>> math.floor(11/1.1)`

10.0

>>> 11//1.1

9.0

The documentation reads "(floored) quotient of x and y". So, why is math.floor(11/1.1) equal to 10, but 11//1.1 equal to 9?

Answer Source

Because 1.1 can't be represented in binary form exactly; the approximation is a littler higher than 1.1 - therefore the division result is a bit too small.

Try the following:

Under Python 2, type at the console:

```
>>> 1.1
1.1000000000000001
```

In Python 3.1, the console will display `1.1`

, but internally, it's still the same number.

But:

```
>>> 11/1.1
10.0
```

As gnibbler points out, this is the result of "internal rounding" within the available precision limits of floats. And as The MYYN points out in his comment, `//`

uses a different algorithm to calculate the floor division result than `math.floor()`

in order to preserve `a == (a//b)*b + a%b`

as well as possible.

Use the `Decimal`

type if you need this precision.