Python3: long int too large to convert to float

Question

I've been checking some topics around here about the same problem I'm getting but they don't seem to help.

My problem is when I try to execute the following code, I get the error found in the title. How do I go around this?

 d=2
 while(n != 1):        
        n = 2
        d = (math.sqrt(2 + d))
        n= (n/d)
        f = (f * (n))
 print (f)

possible duplicate of [OverflowError: long int too large to convert to float in python](http://stackoverflow.com/questions/16174399/overflowerror-long-int-too-large-to-convert-to-float-in-python) — Alex, Mar 09 '14 at 06:15
If you execute that code, you get "NameError: name 'n' is not defined". What code are you actually executing? — Mark Dickinson, Mar 09 '14 at 08:10

Ramchandra Apte · Answer 1 · 2014-03-09T06:20:19.767

3

That's because math.sqrt, as a consequence of using the C sqrt function, works on floating point number which are not unlimited in size. Python is unable to convert the long integer into a floating point number because it is to big. See this question on ways to square root large integers.

Better, you could use the decimal module, which is an unlimited size number type stored in base-10. Use decimal.Decimal(number).sqrt() to find the sqrt of a number.

edited Mar 09 '14 at 06:20

answered Mar 09 '14 at 06:15

Ramchandra Apte

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But, how do I precise exactly that I want my number to 2 d.p? – Sab ಠ_ಠ Mar 09 '14 at 06:22
@Sab I don't understand. What does d.p. stand for? Decimal place? By default, decimal has 38 decimal places so the sqrt will be accurate upto 38 decimal places. – Ramchandra Apte Mar 09 '14 at 06:25
Ya, decimal places. Let's say math.srqt(2) gives a long number(it's irrational), therefore I get the error. So how do I use the decimal module to make it calculate to only 2decimal places, thus avoiding the error? – Sab ಠ_ಠ Mar 09 '14 at 06:28
My answer has the proper code using decimal; no need to use that workaround. – Ramchandra Apte Mar 09 '14 at 06:29
I'm having an error though. [] – Sab ಠ_ಠ Mar 09 '14 at 07:04
1

@Sab Looks like I was wrong about decimal; please use the methods in the [this question](https://stackoverflow.com/questions/15390807/integer-square-root-in-python). (the link is also in the answer). – Ramchandra Apte Mar 09 '14 at 07:22
What about cube root? – the_prole Mar 29 '21 at 02:36

Python3: long int too large to convert to float

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