how to make a code object a function

[Diez B. Roggisch]

Hi,

I've got code objects that are created from strings at runtime that look
like this:

def p_op1(_, args):
_.add_dependenc y('p_op1')

As you migth guess, p_op1 is supposed to become an instance method. Now how
do I make this code object a function-object that can be set into the
instances __dict__ to make it a callable method?

Thanks for any advice,

Diez B. Roggisch

[Peter Otten]

Re: how to make a code object a function

Diez B. Roggisch wrote:
[color=blue]
> Hi,
>
> I've got code objects that are created from strings at runtime that look
> like this:
>
> def p_op1(_, args):
> _.add_dependenc y('p_op1')
>
> As you migth guess, p_op1 is supposed to become an instance method. Now
> how do I make this code object a function-object that can be set into the
> instances __dict__ to make it a callable method?[/color]

Too lazy to type self?
[color=blue][color=green][color=darkred]
>>> import types
>>> class T(object):[/color][/color][/color]
.... def add_dependency( self, name):
.... print "adding dependency", name
....[color=blue][color=green][color=darkred]
>>> t = T()
>>> def p_op1(_, args):[/color][/color][/color]
.... _.add_dependenc y("p_op1")
....[color=blue][color=green][color=darkred]
>>> t.p_op1 = types.MethodTyp e(p_op1, t)
>>> t.p_op1(1)[/color][/color][/color]
adding dependency p_op1[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

By the way, types.MethodTyp e and new.instancemet hod seem to be equivalent.

Peter

[Diez B. Roggisch]

Re: how to make a code object a function

Hi,
[color=blue]
> Too lazy to type self?[/color]

yes. I used an underscore in java as indication for member-variables, so
typing _. appeared natural to me.
[color=blue][color=green][color=darkred]
>>>> import types
>>>> class T(object):[/color][/color]
> ... def add_dependency( self, name):
> ... print "adding dependency", name
> ...[color=green][color=darkred]
>>>> t = T()
>>>> def p_op1(_, args):[/color][/color]
> ... _.add_dependenc y("p_op1")
> ...[color=green][color=darkred]
>>>> t.p_op1 = types.MethodTyp e(p_op1, t)
>>>> t.p_op1(1)[/color][/color]
> adding dependency p_op1[color=green][color=darkred]
>>>>[/color][/color]
>
> By the way, types.MethodTyp e and new.instancemet hod seem to be equivalent.[/color]

Interesting - from the docs, I can't see that types.MethodTyp e has a
constructor like this.

Your example above gave me the idea to simply exec() my code-object, so in
the end I have a new function called p_op1 in my scope - that function I
can get using locals()['p_op1']. I know the name, so that works.

Just out of curiosity - is there a way to know the name of a code object you
know nothing about except that it will become a function definition? I
guess I could go for some AST-stuff looking for a "def foo" statement, so I
know I will end up having defined foo when exec'ing the code object.

Regards,

Diez B. Roggisch

[Emile van Sebille]

Re: how to make a code object a function

Diez B. Roggisch:[color=blue]
> Just out of curiosity - is there a way to know the name of a code[/color]
object you[color=blue]
> know nothing about except that it will become a function definition?[/color]
I[color=blue]
> guess I could go for some AST-stuff looking for a "def foo"[/color]
statement, so I[color=blue]
> know I will end up having defined foo when exec'ing the code object.
>[/color]

I'm not quit sure what you're asking, but does sets help?
[color=blue][color=green][color=darkred]
>>> ns = locals().keys()
>>> testing = 1
>>> print sets.Set(locals ().keys()) - sets.Set(ns)[/color][/color][/color]
Set(['testing'])


Emile van Sebille
emile@fenx.com

[Peter Otten]

Re: how to make a code object a function

Diez B. Roggisch wrote:
[color=blue]
> Interesting - from the docs, I can't see that types.MethodTyp e has a
> constructor like this.[/color]

It get's really funny if you look into the source (types.py):

class _C:
def _m(self): pass
ClassType = type(_C)
UnboundMethodTy pe = type(_C._m) # Same as MethodType
_x = _C()
InstanceType = type(_x)
MethodType = type(_x._m)

Seems we caught them cheating here :-)
[color=blue]
> Just out of curiosity - is there a way to know the name of a code object
> you know nothing about except that it will become a function definition? I
> guess I could go for some AST-stuff looking for a "def foo" statement, so
> I know I will end up having defined foo when exec'ing the code object.[/color]

You could provide a separate namespace and then extract only the callables:
[color=blue][color=green][color=darkred]
>>> d = {}
>>> exec "factor=2\n def alpha(s, t): print factor*t" in d
>>> d.keys()[/color][/color][/color]
['__builtins__', 'alpha', 'factor'][color=blue][color=green][color=darkred]
>>> funcs = dict([(n,f) for n, f in d.iteritems() if callable(f)])
>>> funcs[/color][/color][/color]
{'alpha': <function alpha at 0x4029025c>}[color=blue][color=green][color=darkred]
>>> funcs["alpha"](None, 2)[/color][/color][/color]
4

Peter

[Diez B. Roggisch]

Re: how to make a code object a function

> You could provide a separate namespace and then extract only the[color=blue]
> callables:
>[color=green][color=darkred]
>>>> d = {}
>>>> exec "factor=2\n def alpha(s, t): print factor*t" in d
>>>> d.keys()[/color][/color]
> ['__builtins__', 'alpha', 'factor'][color=green][color=darkred]
>>>> funcs = dict([(n,f) for n, f in d.iteritems() if callable(f)])
>>>> funcs[/color][/color]
> {'alpha': <function alpha at 0x4029025c>}[color=green][color=darkred]
>>>> funcs["alpha"](None, 2)[/color][/color]
> 4[/color]

And again I learned something completely new - cool. Thanks,

Diez