Can't extend function type

[Diez B. Roggisch]

Paul Rubin wrote:
> Oh well.  I had wanted to be able to define two functions f and g, and
> have f*g be the composition of f and g.
> 
>     >>> func_type = type(lambda: None)
>     >>> class composable_function(func_type):
>     ...   def __mult__(f,g):
>     ...     def c(*args, **kw):
>     ...       return f(g(*args, **kw))
>     ...     return c
>     ...
>     Traceback (most recent call last):
>       File "<stdin>", line 1, in ?
>     TypeError: Error when calling the metaclass bases
>         type 'function' is not an acceptable base type
>     >>>
> 
> Seems like a wart to me.

Well - function inheritance is not known so far in python - and in no 
other language I know.

How do you expect to create f and g, even if above construct would work? 
Basdically you want __mult__ being part of f or g when python encounters 
  something like this

f * g

But then how did you plan to declare f?

def f(composable_function)(x):
     pass


obviously won't work.

So the only way to achieve this with current semantics is to make f anf 
g objects with a call methods. In that very moment, you're done - as 
extending from object is no problem :)


class ComposeableFunction(object):

     def __call__(self, *args, **kwargs):
	return self.second(self.first(*args, **kwargs))

     def __mul__(self, other):
	nc = ComposeableFunction()
	nc.first = other
	nc.second = self
	return nc


class f(ComposeableFunction):
     def __call__(self, x):
	return x * 2


class g(ComposeableFunction):
     def __call__(self, x):
	return x + 2


f = f()
g = g()

print f(4)
print g(4)
print (f*g)(4)


Diez