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[MDX] Dream code for CustomVertex

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I want that in the code of creation and initialisation of VertexBuffer the type's structure of the CustomVertex was not duplicated! Code i want to refactoring:
public override void InitGraphVB( Device ADevice )
{
	int count = 100;
	System.Type vertType = typeof( CustomVertex.PositionNormalColored );

	VertexFormats vertFormat = CustomVertex.PositionNormalColored.Format;

			
	VertexBuffer vb = new VertexBuffer( vertType, count, ADevice, Usage.WriteOnly, vertFormat, Pool.Managed );

	CustomVertex.PositionNormalColored[] verts = ( CustomVertex.PositionNormalColored[] ) vb.Lock( 0, 0 ); 
	
	for ( int i = 0; i < count; i++ )
	{	
		Vector3 pos = ...;
		Vector3 norm = ...;
		System.Drawing.Color color = ...;
		verts[ i ] = new CustomVertex.PositionNormalColored( pos, norm, color );
	}
	vb.Unlock();
}


Dream code to illustrate the idea:
public override void InitGraphVB( Device ADevice )
{
	int count = 100;
	System.Type vertType = typeof( CustomVertex.PositionNormalColored );

	VertexFormats vertFormat = ( VertexFormats ) ( classof( vertType ).GetField( "Format" ) );

			
	VertexBuffer vb = new VertexBuffer( vertType, count, ADevice, Usage.WriteOnly, vertFormat, Pool.Managed );

	classof( vertType )[] verts = ( classof( vertType )[] ) vb.Lock( 0, 0 ); 

	for ( int i = 0; i < count ; i++ )
	{	
		Vector3 pos = ...;
		Vector3 norm = ...;
		System.Drawing.Color color = ...;

		object[] constrParams = { ( object )pos, ( object )norm , ( object )color };

		verts[ i ] = Activator.CreateInstance( vertType, constrParams );
	}
	vb.Unlock();
}


The main difficulty for me how get class by his type: classof( vertType )? Also i don't know how get access to class members knowing his name like this: classof( vertType ).GetField( "Format" )? [Edited by - Maxim Skachkov on April 4, 2006 2:06:01 PM]

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Hi,duhroach!
No problem in the Constructor(). The constructors for a vertex buffer are identical for both samples of the code.

I mean the "CustomVertex.PositionNormalColored" code fragment should appeared only once.

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