| How to access a siblings config params? [message #3945] |
Mon, 15 June 2009 14:27  |
Eclipse User |
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Champs,
I am trying to access a module's config param from the module$use() function of another module in the same package. I'm currently using the following:
function module$use()
{
var Settings = this.$package.Settings;
if (Settings.ipc...) {
:
}
}
But in the documentation for the XDCscript Language Summary it does not recommend using this method.
pkg.ModName Module
A particular module contained within this package, unless ModName matches a name in the xdc.IPackage interface -- in this case the xdc.IPackage interface value is returned. Because of this ambiguity, this method of referencing a package's modules is deprecated.
What is the recommended way to access a siblings config params? Should I be using the following:
var Settings = xdc.useModule(this.$package.$name + '.Settings');
Or is there a better way?
Thanks
~ Ramsey
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| Re: How to access a siblings config params? [message #3952 is a reply to message #3945] |
Mon, 15 June 2009 15:20  |
| Eclipse User |
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Ramsey Harris wrote:
> Champs,
>
> I am trying to access a module's config param from the module$use()
> function of another module in the same package. I'm currently using the
> following:
>
> function module$use()
> {
> var Settings = this.$package.Settings;
> if (Settings.ipc...) {
> :
> }
> }
>
> But in the documentation for the XDCscript Language Summary it does not
> recommend using this method.
>
> pkg.ModName Module
>
> A particular module contained within this package, unless ModName
> matches a name in the xdc.IPackage interface -- in this case the
> xdc.IPackage interface value is returned. Because of this ambiguity,
> this method of referencing a package's modules is deprecated.
>
For what it's worth: if you follow the module naming conventions, there
will not be a name conflict.
> What is the recommended way to access a siblings config params? Should I
> be using the following:
>
> var Settings = xdc.useModule(this.$package.$name + '.Settings');
>
> Or is there a better way?
You can also use just xdc.module(). xdc.useModule() implies that the
named module will be initialized at runtime (if unless it's metaonly),
xdc.module() does not force the module to be part of the application.
>
> Thanks
> ~ Ramsey
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