Why does 1**2**3**4**5 raise a MemoryError?

Steven D'Aprano steve+comp.lang.python at pearwood.info
Sun Mar 31 09:33:32 CEST 2013


On Sat, 30 Mar 2013 23:56:46 -0700, morphex wrote:

> Hi.
> 
> I was just doodling around with the python interpreter today, and here
> is the dump from the terminal:
> 
> morphex at laptop:~$ python
> Python 2.7.3 (default, Sep 26 2012, 21:53:58) [GCC 4.7.2] on linux2
> Type "help", "copyright", "credits" or "license" for more information.
>>>> 1**2
> 1
>>>> 1**2**3
> 1
>>>> 1**2**3**4
> 1L
>>>> 1**2**3**4**5
> Traceback (most recent call last):
>   File "<stdin>", line 1, in <module>
> MemoryError
>>>> 
>>>> 
> Does anyone know why this raises a MemoryError?  Doesn't make sense to
> me.

Because exponentiation is right-associative, not left.

1**2**3**4**5 is calculated like this:

1**2**3**4**5
=> 1**2**3**1024
=> 1**2**373...481  #  489-digit number
=> 1**(something absolutely humongous)
=> 1

except of course you get a MemoryError in calculating the intermediate 
values.

In other words, unlike you or me, Python is not smart enough to realise 
that 1**(...) is automatically 1, it tries to calculate the humongous 
intermediate result, and that's what fails.

For what it's worth, that last intermediate result (two to the power of 
the 489-digit number) has approximately a billion trillion trillion 
trillion trillion trillion trillion trillion trillion trillion trillion 
trillion trillion trillion trillion trillion trillion trillion trillion 
trillion trillion trillion trillion trillion trillion trillion trillion 
trillion trillion trillion trillion trillion trillion trillion trillion 
trillion trillion trillion trillion trillion trillion digits.

(American billion and trillion, 10**9 and 10**12 respectively.)



-- 
Steven



More information about the Python-list mailing list