Yeah. This was on 0.4, and it sounds like the change does what I was after. 
Thanks!

--Tim

On Saturday, January 30, 2016 08:39:00 AM Michael Hatherly wrote:
> Tim, is this with 0.4 or 0.5? The behaviour of type/constructor docs was
> changed quite recently
> to be more similar to that of function/method docs. On 0.5 `help?> Foo`
> should concatenate all
> the docstrings for `Foo` and it's constructors. Is that the behaviour
> you're looking for?
> 
> -- Mike
> 
> On Saturday, 30 January 2016 14:26:25 UTC+2, Tim Holy wrote:
> > Do we have an easy way to retrieve documentation for constructors
> > separately
> > from the documentation of a type? I spent a few minutes looking at the
> > source
> > and searching issues, and got to the point where I thought it was better
> > to
> > ask before, say, considering a PR.
> > 
> > Illustration:
> >     """
> >     Foo is a type that represents some amazing stuff
> >     """
> >     type Foo
> >     end
> >     
> >     """
> >     `foo = Foo(7)` creates an empty `Foo` instance with room for 7
> > 
> > LittleFoos
> > in it.
> > 
> >     """
> >     Foo(n::Integer) = nothing
> >     
> >     """
> >     `foo = Foo(x, y)` puts the LittleFoos `x` and `y` into a grown-up Foo.
> >     """
> >     Foo(x, y) = nothing
> > 
> > Now let's try it:
> >     help?> Foo
> >     search: Foo floor ifloor pointer_from_objref OverflowError
> > 
> > RoundFromZero
> > FileMonitor functionloc functionlocs StackOverflowError Factorization
> > OutOfMemoryError
> > 
> >       Foo is a type that represents some amazing stuff
> > 
> > Now, if I know how to call the constructor I want, then it's no problem to
> > 
> > retrieve the documentation:
> >     help?> Foo(7)
> >     
> >       foo = Foo(7) creates an empty Foo instance with room for 7
> > 
> > LittleFoos in
> > it.
> > 
> > But what if I don't know my choices? Here are two things I tried:
> >     julia> @eval @doc $(methods(Foo))
> >     
> >     help?> call(Type{Foo})
> >   
> >   call(x, args...)
> >   
> >   If x is not a Function, then x(args...) is equivalent to call(x,
> > 
> > args...).
> > This means that function-like behavior can be added to any type by
> > defining new
> > call methods.
> > 
> > Neither produced useful results. Any thoughts?
> > 
> > Best,
> > --Tim

Reply via email to