Re: [xsl] XSLT 3.0: What is the default visibility of a package component inside the declaring package ?

Subject: Re: [xsl] XSLT 3.0: What is the default visibility of a package component inside the declaring package ?
From: "Dimitre Novatchev dnovatchev@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Sun, 2 Nov 2014 17:23:58 -0000
My question was probably incorrect -- the "visibility" attribute
defines the visibility of a component in a using package, not in the
package itself.

And the Spec contains a 5-step explanation how to determine the
visibility of a component. If there is no visibility attribute on the
component and on an <xsl:expose> element, then the visivility of the
component is "private" -- which makes sense and is convenient.

Cheers,
Dimitre

On Sun, Nov 2, 2014 at 9:10 AM, Dimitre Novatchev dnovatchev@xxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
> What is the default visibility of a package component inside the
> declaring package, if the component has no "visibility" attribute
> specified on its declaration?
>
> I believe that the default visibility should be "public", however I
> couldn't find any statement about this in the latest XSLT 3.0
> specification.
>
> So, could someone confirm my guess, and/or point the text in the Spec
> that explains this?
>
> If there isn't really such definition, I believe it needs to be added
> to the specification.
>
> --
> Cheers,
> Dimitre Novatchev
> 



-- 
Cheers,
Dimitre Novatchev
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