getattr(foo, 'foobar') not the same as foo.foobar?

[castironpi at gmail.com]

On Mar 13, 8:15 pm, castiro... at gmail.com wrote:
> On Mar 13, 7:45 pm, castiro... at gmail.com wrote:
>
>
>
>
>
> > On Mar 13, 7:18 pm, Mel <mwil... at the-wire.com> wrote:
>
> > > Diez B. Roggisch wrote:
> > > >> My understanding is that foo.bar does *not* create a new object.
>
> > > > Your understanding is not correct.
>
> > > >>  All it
> > > >> does is return the value of the bar attribute of object foo.  What new
> > > >> object is being created?
>
> > > > A bound method. This happens through the descriptor-protocol. Please see
> > > > this example:
>
> > > > class Foo(object):
> > > >     def bar(self):
> > > >         pass
>
> > > > f = Foo()
> > > > a = Foo.bar
> > > > b = f.bar
> > > > c = f.bar
>
> > > > print a, b, c
> > > > print id(b), id(c)
>
> > > (What Diez said.)  From what I've seen, f.bar creates a bound method
> > > object by taking the unbound method Foo.bar and binding its first
> > > parameter with f.  This is a run-time operation because it's easy to
> > > re-assign some other function to the name Foo.bar, and if you do, the
> > > behaviour of f.bar() will change accordingly.
>
> > > You can get some very useful effects from these kinds of games.  You
> > > can make f into a file-like object, for example, with
>
> > > import sys
> > > f.write = sys.stdout.write
>
> > > Here, f.write *is* a straight attribute of f, although it's a built-in
> > > method of the file class.  It's still bound, in a way, to sys.stdout.
> > >   I'm assuming that a different example could create an attribute of f
> > > that's a bound method of some other object entirely.  I've verified
> > > that f.write('howdy') prints 'howdy' on standard output.
>
> > Accordingly,
>
> > f.write= types.MethodType( sys.stdout.__class__.write, sys.stdout ).
>
> Ah.  Because this doesn't.
> That is, because sys.stdout.write is -not- a user-defined function.
> What it is, is a bound member function, and only the former is
> converted/wrapped/bound*, as it is in the subsequent example.

Last thing, sorry:  Does MethodType.__get__ just return self, or is it
not called, due to some type checking?