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

[castironpi at gmail.com]

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 ).

It depends on what you want the implicit first (self) to be-- f or
sys.stdout.

But how come this works?

>>> import types
>>> import sys
>>>
>>> class C:
...     write= sys.stdout.write
...     def g( self ):
...             self.write( 'was in \'g\'\n' )
...
>>> c= C()
>>> c.g()
was in 'g'

Shouldn't 'write' be getting extra parameters?  Sounds fishy, not to
mix metaphors.