On Mar 13, 7:18 pm, Mel <[EMAIL PROTECTED]> 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.
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