Importing Custom namespace

[kerry_w_brown]

I'd like to know the logic AutoCAD used to handle loading Custom Assemblies from a namespace using directive.

Are all types defined in a namespace imported ?
. . . .  or only the types required by that particular assembly ?

Does 'AutoCAD' check if a namespace has been previously loaded, and only load if not previously loaded. ?

If so, can we assume that several custom assemblies can/do use the same loaded 'namespace' for the types they have in common. ??

Regards,
Kerry

[_gile]

Hi Kerry,

This is not specific to AutoCAD, this is the way Visual Studio works.

From Microsoft documentation:

" the using directive imports all the types from a single namespace"

except with the 'static' modifier:

"The static modifier imports the static members and nested types from a single type rather than importing all the types in a namespace."

I do not think the word "load" is appropriated for namespaces. Namespaces are related to the type names, not their implementation.

[_gile]

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@kerry_w_brown  a écrit :

can we assume that several custom assemblies can/do use the same loaded 'namespace' for the types they have in common. ??

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I'm not sure to fully understand.

If several assemblies can use the same namespace (Autodesk.AutoCAD.Runtime is used by the accoremgd, acdbmgd and acmgd assemblies), there can't be the same type name for different types in different assemblies.

[kerry_w_brown]

Hi Gilles,

I expressed that badly.
I want to be sure that several assemblies will be able to share stand alone libraries.

[daniel_cadext]

Not sure if this helps, but

Reflect on Autodesk.AutoCAD.Runtime.ExtensionLoader, OnAssemblyResolve, OnAssemblyLoad, ProcessAssembly

There’s a hashmap of loaded assemblies (and their references) and of classes that have been declared with one of AutoCAD’s custom attributes.

Guessing that AutoCAD knows what assemblies have been loaded, directly, or by reference, and if a type is loaded that has an attribute.

Everything else is probably done by the framework.