Confusing math problem

Oscar Benjamin oscar.j.benjamin at
Fri Feb 22 01:04:30 CET 2013

On 21 February 2013 23:41, Schizoid Man <schiz_man at> wrote:
> "Oscar Benjamin" <oscar.j.benjamin at> wrote in
>> Then you want operator.pow:
>>>>> import operator
>>>>> operator.pow(3, 2)
>> 9
>> math.pow is basically the (double)pow(double, double) function from
>> the underlying C library. operator.pow(a, b) is precisely the same as
>> a**b.
> So how is operator.pow() different from just pow()?

operator.pow(a, b) calls a.__pow__(b). This is also what a**b does. If
a and b are ints then the __pow__ method will create the appropriate
int exactly without overflow and return it. You can make your own
class and give it a __pow__ function that does whatever you like:

>>> class Number(object):
...   def __pow__(self, other):
...     print('__pow__ called, returning 3')
...     return 3
>>> n = Number()
>>> result = n**3
__pow__ called, returning 3
>>> result

operator.pow will call the method:

>>> import operator
>>> operator.pow(n, 3)
__pow__ called, returning 3

math.pow will not:

>>> import math
>>> math.pow(n, 3)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: a float is required

The math module essentially exposes the functions that would be
available in C after importing math.h. So when you call math.pow(a,
b), a and b are if possible converted to float (which corresponds to a
double in C) and then the result is computed and returned as a float.


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