Aivar - 1 month ago 5x

Python Question

I know that most decimals don't have an exact floating point representation (Is floating point math broken?).

But I don't see why

`4*0.1`

`0.4`

`3*0.1`

both values actually have ugly decimal representations:

`>>> 3*0.1`

0.30000000000000004

>>> 4*0.1

0.4

>>> from decimal import Decimal

>>> Decimal(3*0.1)

Decimal('0.3000000000000000444089209850062616169452667236328125')

>>> Decimal(4*0.1)

Decimal('0.40000000000000002220446049250313080847263336181640625')

Answer

The simple answer is because `3*0.1 != 0.3`

due to quantization (roundoff) error (whereas `4*0.1 == 0.4`

because multiplying by a power of two is usually an "exact" operation).

You can use the `.hex`

method in Python to view the internal representation of a number (basically, the *exact* binary floating point value, rather than the base-10 approximation). This can help to explain what's going on under the hood.

```
>>> (0.1).hex()
'0x1.999999999999ap-4'
>>> (0.3).hex()
'0x1.3333333333333p-2'
>>> (0.1*3).hex()
'0x1.3333333333334p-2'
>>> (0.4).hex()
'0x1.999999999999ap-2'
>>> (0.1*4).hex()
'0x1.999999999999ap-2'
```

0.1 is 0x1.999999999999a times 2^-4. The "a" at the end means the digit 10 - in other words, 0.1 in binary floating point is *very slightly* larger than the "exact" value of 0.1 (because the final 0x0.99 is rounded up to 0x0.a). When you multiply this by 4, a power of two, the exponent shifts up (from 2^-4 to 2^-2) but the number is otherwise unchanged, so `4*0.1 == 0.4`

.

However, when you multiply by 3, the little tiny difference between 0x0.99 and 0x0.a0 (0x0.07) magnifies into a 0x0.15 error, which shows up as a one-digit error in the last position. This causes 0.1*3 to be *very slightly* larger than the rounded value of 0.3.

Python 3's float `repr`

is designed to be *round-trippable*, that is, the value shown should be exactly convertible into the original value. Therefore, it cannot display `0.3`

and `0.1*3`

exactly the same way, or the two *different* numbers would end up the same after round-tripping. Consequently, Python 3's `repr`

engine chooses to display one with a slight apparent error.

Source (Stackoverflow)

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