How can Python represent 2**500 as an int?

Question

What type of int is the result of this calculation? Why isn't there a numerical overflow issue?

>>> 2**500
3273390607896141870013189696827599152216642046043064789483291368096133796404674554883270092325904157150886684127560071009217256545885393053328527589376
>>> type(2**500)
<class 'int'>

Because python `int` objects are arbitrarily sized, as long as you have memory and address space, you are good — juanpa.arrivillaga, May 21 '21 at 13:49
Don't confuse Python's `int` type with whatever integer types your hardware supports. A Python `int` is really like a resizable array of base-2**30 digits (or something like that). — chepner, May 21 '21 at 13:51
@CharlesDuffy certainly I don't think it's unique to 2**500. In fact I experimented with `2**501` and the result was similar — Alexander Soare, May 21 '21 at 13:51
Can you clarify your question? The type of the calculation is... ``int``. As your example shows. — MisterMiyagi, May 21 '21 at 13:53
@MisterMiyagi I believe the "Why isn't there a numerical overflow issue?" says it all Hence the accepted answer — Alexander Soare, May 21 '21 at 13:54
Does this answer your question? [How Python is working with a number bigger than the 64-bit unsigned integer limit?](https://stackoverflow.com/questions/63922356/how-python-is-working-with-a-number-bigger-than-the-64-bit-unsigned-integer-limi) — MisterMiyagi, May 21 '21 at 13:57

score 1 · Answer 1 · answered May 21 '21 at 13:51

1

I think this is implemented as a bignum which can handle arbitrarily large integer. You can find a more detailed discussion here, Handling very large numbers in Python

answered May 21 '21 at 13:51

EyeOfAgamotto

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score 1 · Accepted Answer · answered May 21 '21 at 13:52

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From this link "In Python, integers are implemented as “long” integer objects of arbitrary size", they do not overflow https://docs.python.org/3/c-api/long.html#integer-objects

answered May 21 '21 at 13:52

Achille G

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How can Python represent 2**500 as an int?

2 Answers2