On Feb 22, 3:18 am, Peter Otten <[EMAIL PROTECTED]> wrote:
> Alejandro Dubrovsky wrote:
> > def autoassign(_init_):
> > import inspect
> > import functools
>
> > argnames, _, _, defaults = inspect.getargspec(_init_)
> > argnames = argnames[1:]
>
> > indentation = ' '
> > settings = ['self.%s = %s' % (arg[1:], arg) for arg in argnames
> > if arg[0] == '_']
>
> > if len(settings) <= 0:
> > return _init_
>
> > if defaults is None:
> > args = argnames[:]
> > else:
> > args = argnames[:-len(defaults)]
> > for key, value in zip(argnames[-len(defaults):],defaults):
> > args.append('%s=%s' % (key, repr(value)))
>
> > template = """def _autoassign(self, %(args)s):
> > %(setting)s
> > _init_(self, %(argnames)s)
> > """ % {'args' : ", ".join(args), 'setting' : "\n".join(['%s%s' %
> > (indentation, setting) for setting
> > in settings]), 'argnames' : ', '.join(argnames)}
>
> > try:
> > exec template
> > except SyntaxError, e:
> > raise SyntaxError('%s. line: %s. offset %s:\n%s' %
> > (e.msg, e.lineno, e.offset, template))
> > return _autoassign
>
>
[snip]
> > Is there a way to bind the _init_ name at exec time?
>
> Use a dedicated namespace:
>
> namespace = dict(_init_=_init_)
> exec template in namespace
> return namespace["_autoassign"]
>
> Peter
That works! Excellent, thanks.
I still don't understand why the original doesn't work. I thought
exec with no namespace specified used the current context, in which
_init_ would be _init_ already. But understanding can wait.
ale
--
http://mail.python.org/mailman/listinfo/python-list
