Is __mul__ sufficient for operator '*'?

Muhammad Alkarouri malkarouri at gmail.com
Mon Oct 19 20:31:44 EDT 2009


Hi everyone,

I was having a go at a simple implementation of Maybe in Python when I
stumbled on a case where x.__mul__(y) is defined while x*y is not.

The class defining x is:

class Maybe(object):
    def __init__(self, obj):
        self.o = obj
    def __repr__(self):
        return 'Maybe(%s)' % object.__getattribute__(self, "o")
    def __getattribute__(self, name):
        try:
            o = object.__getattribute__(self, "o")
            r = getattr(o,name)
            if callable(r):
                f = lambda *x:Maybe(r(*x))
                return f
            else:
                return Maybe(r)
        except:
            return Maybe(None)

The code exercising this class is:

>>> x=Maybe(9)
>>> x.__mul__(7)
Maybe(63)
>>> x*7

Traceback (most recent call last):
  File "<pyshell#83>", line 1, in <module>
    x*7
TypeError: unsupported operand type(s) for *: 'Maybe' and 'int'

The farthest I can go in this is that I presume that __mul__ (as
called by operator *) is supposed to be a bound method while I am
returning a lambda function. Is this correct? And How can I make the
implementation support such operators?

Cheers,

Muhammad Alkarouri



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