Hey Simon and Darij,
Doing :meth:`Class.foo()` will properly link if and only if the class is
imported into the global namespace. At least that has been my experience.
IMO it should find a class matching it through all linked documentation
starting from the current location, irregardless if it is imported or not...
(I think the doc for sphinx says it does this, but it never seems to work
for me.)
Best,
Travis
On Saturday, June 29, 2013 11:35:05 AM UTC+2, Simon King wrote:
>
> Hi Darij,
>
> On 2013-06-29, Darij Grinberg <darijg...@gmail.com <javascript:>> wrote:
> > thanks. This comes too late for my patch, but I'll think about it in
> > the future (I didn't think about splitting the summary into a one-line
> > and a detailed part).
>
> If I recall correctly, each doc string is supposed to be of this form. I
> don't know if this is written in the developer guide, though.
>
> > BTW, is there a way to refer to
> >:meth:`SuperClass.foo` without giving the whole path
> > ("sage.combinat......") to SuperClass?
>
> If SuperClass is defined in the same Python module, then
> :meth:`SuperClass.foo` is enough. If it is in a different Python module,
> I think you need to provide the whole path. But it is possible that the
> resulting link in the documentation is abbreviated. Namely, if you type
> :meth:`~sage.combinat.whatever_module.SuperClass.foo`, then the link in
> the documentation will only appear as "foo()". Hence, if you prefer to
> give it in this form, it might be good to *additionally* provide a link
> to the class, hence, something like
>
> NOTE:
>
> This overloads :meth:`~sage.combinat.whatever_module.SuperClass.foo`
> of :class:`~sage.combinat.whatever_module.SuperClass`, for
> efficiency reasons.
>
> Best regards,
> Simon
>
>
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