Muhammad Alkarouri wrote:
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 product 7*x will execute __mul__, but to catch x*7, you need to
define __rmul__ (the 'r' stands for reverse or some such).
However, in fact, you have *not* defined __mul__ in your class. Your
__getattribute__ is catching __mul__ as an undefined reference, and
doing something with it -- not sure what though.
As proof, continue testing: x.__mul__ and x.__rmul__ both return
values (lambdas defined within __getattribute__) and neither x*7 or 7*x
work.
If you want both x* and 7*x to be defined, try
def __mul__(self,r):
...
def __rmul__(self,r):
....
Or the operation is commutative, perhaps you can get away with reusing
__mul__ for both.
def __mul__(self,r):
...
__rmul__ = __mull__
Gary Herron
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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