Last decimals are random counting float

Question

I am starting with learning Python. I currently have the following small piece of code:

a = float(input())
b = float(3)
while a < b:
    print(a + 0.0001)
    a+=0.0001

This counts, but there appear decimals in the end of the float. Why do they come there and what are they?

Output:

enter image description here

What makes you think they're random? Also, why include a *picture* of your output? Read [this](http://stackoverflow.com/q/1089018/3001761) or [this](http://stackoverflow.com/q/2986150/3001761) or [this](http://stackoverflow.com/q/1594985/3001761) or ... — jonrsharpe, Jul 06 '14 at 19:41
[What Every Computer Scientist Should Know About Floating-Point Arithmetic](http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html) — The Paramagnetic Croissant, Jul 06 '14 at 19:45
It makes me sad that the accepted answer at the dupe Q is so poor — David Heffernan, Jul 07 '14 at 06:31

David Heffernan · Answer 1 · 2014-07-07T06:50:21.813

1

This is because the numbers in the question are not exactly represented by binary floating point. Start with 0.0001. The closest double precision value is in fact:

0.00010000000000000000479217360238592959831294137984514236450195312 5

So when you write 0.0001 what you actually get is the number above. That is then added to whatever value you input, and then the result is the closest representable value to that.

If this comes as a shock to you, then I suggest that you read David Goldberg's timeless article What Every Computer Scientist Should Know About Floating-Point Arithmetic.

edited Jul 07 '14 at 06:50

answered Jul 06 '14 at 19:55

David Heffernan

601,492
42
1,072
1,490

Can we have a few more decimal places in the value for `0.0001`, please? (I get `0.000100000000000000004792173602385929598312941379845142364501953125`.) – Mark Dickinson Jul 07 '14 at 06:41
@mark you are correct. I mis-copied the value from rob's page – David Heffernan Jul 07 '14 at 06:50

Last decimals are random counting float

1 Answers1