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# Batching Objects together

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I have a quick little C# app that draws 3D objects (primitives)

I have 3 objects that I want to draw:

Cubes
Cones
Spheres

At the moment I am drawing them as they come in. So I could have 3 cones, 1 Cube, 2 spheres, 1 cone, 5 cubes etc.

As you can imagine this is in-efficient. Basically what I would like to do is batch the objects together so that all like objects are together eg: all cones, all cubes, all spheres are together.

Just before I finish a frame I will refer to the batch and draw the primitives.

What I was planning was to create a simple design pattern that simply added the objects to the relevant structure (either an array list or perhaps a hash table.) Then iterate through the structures when ready.

Anyone have any ideas on how to organise or store basic objects in a structure that would all for the most efficient insertion and retrieval?

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Commented:
>> As you can imagine this is in-efficient.
Why?

Drawing objects is 3D space is usually done by z-order, and not by their types. However, you can keep all your objects in 3 lists, every list contains objects of one type.

Commented:
In this situation Z-Order isnt important, but rather squezzing that little bit extra out of it. I could pipe all objects into their various lists and pull them out as needed.
Commented:
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Commented:
I think it probably doesn't matter. With 3d graphics, almost all the processing time is in the transforms, etc that go into processing the objects. Optimizing the storage is like taking an eyedropper of water out of niagra falls. The fastest order to process them is the one that eliminates the most transforming, lighting, texturing, and rendering. As someone said, that's usually done by descending z-order, but there are more complictated algorithms (BSP-trees come to mind) that are even better.

If really all you care about is the storage for some reason, pallosp is right - use 3 queues. If you need a single data structure to retrieve from, wrap the 3 queues in a simple class that inserts/retrieves from the right one.
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