Why does 1**2**3**4**5 raise a MemoryError?
Steven D'Aprano
steve+comp.lang.python at pearwood.info
Sun Mar 31 03:33:32 EDT 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
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