Rahul wrote: > Consider the following: > def a(x): > return x+1 > > def b(f): > def g(*args,**kwargs): > for arg in args: > print arg > return f(*args,**kwargs) > return g > > a.__call__ = b(a.__call__) > > now calling a(1) and a.__call__(1) yield 2 different results!! > i.e. for functions a(1) doesnt seem to be translated to a.__call__ if > you assign a new value to a.__call__?
I don't know why this happens, but setting the __call__ attribute of a is a pretty strange thing to do. Why not just set a instead? The original function a(x) will still be stored as a closure in what is returned from b(). If this is of purely academic interest then the answer is I don't know. :) -- Michael Hoffman -- http://mail.python.org/mailman/listinfo/python-list
