Function argument conformity check
george.sakkis at gmail.com
Wed Jun 18 23:05:48 CEST 2008
On Jun 18, 3:41 pm, dlists.... at gmail.com wrote:
> On Jun 18, 3:13 pm, Cédric Lucantis <o... at no-log.org> wrote:
> > Hi,
> > Le Wednesday 18 June 2008 20:19:12 dlists.... at gmail.com, vous avez écrit :
> > > Hi. I am looking for a way to check if some given set of (*args,
> > > **kwds) conforms to the argument specification of a given function,
> > > without calling that function.
> > > For example, given the function foo:
> > > def foo(a, b, c): pass
> > > and some tuple args and some dict kwds, is there a way to tell if i
> > > _could_ call foo(*args, **kwds) without getting an exception for those
> > > arguments? I am hoping there is a way to do this without actually
> > > writing out the argument logic python uses.
> > Each function object is associated to a code object which you can get with
> > foo.func_code. Two of this object's attributes will help you: co_argcount and
> > co_varnames. The first is the number of arguments of the function, and the
> > second a list of all the local variables names, including the arguments
> > (which are always the first items of the list). When some arguments have
> > default values, they are stored in foo.func_defaults (and these arguments are
> > always after non-default args in the co_argnames list).
> > Finally, it seems that some flags are set in code.co_flags if the function
> > accepts varargs like *args, **kwargs, but I don't know where these are
> > defined.
> > Note that I never found any doc about that and merely guessed it by playing
> > with func objects, so consider all this possibly wrong or subject to change.
> > --
> > Cédric Lucantis
> I am aware of these attributes, although you can get them all in a
> more organized form using the getfullargspec function in the inspect
> module from the standard library.
> The problem is that using these attributes, I would essentially have
> to re-write the logic python uses when calling a function with a given
> set of arguments. I was hoping there is a way to get at that logic
> without rewriting it.
Sure; copy it from someone that has already done it ;-)
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