我们接着系列二的问题继续讲解,系列二中的问题是如果多个材质是相同的,它没有去优化,接下来我们将其优化一下,首先要找到在哪里去优化,我们看下面的for循环语句也是系列二的代码,如下:
for (int s = 0; s < meshFilter.sharedMesh.subMeshCount; s++) { int materialArrayIndex = 0; for (materialArrayIndex = 0; materialArrayIndex < materials.Count; materialArrayIndex++) { if (materials[materialArrayIndex] == meshRenderer.sharedMaterials[s]) break; } if (materialArrayIndex == materials.Count) { materials.Add(meshRenderer.sharedMaterials[s]); combineInstanceArrays.Add(new ArrayList()); } CombineInstance combineInstance = new CombineInstance(); combineInstance.transform = meshRenderer.transform.localToWorldMatrix; combineInstance.subMeshIndex = s; combineInstance.mesh = meshFilter.sharedMesh; (combineInstanceArrays[materialArrayIndex] as ArrayList).Add(combineInstance); }
这个代码里它没有判断是否有相同的材质,接下来我们将其重新修改一下:
for (int s = 0; s < meshFilter.sharedMesh.subMeshCount; s++) { int materialArrayIndex = Contains(materials, meshRenderer.sharedMaterials[s].name); if (materialArrayIndex == -1) { materials.Add(meshRenderer.sharedMaterials[s]); materialArrayIndex = materials.Count - 1; } combineInstanceArrays.Add(new ArrayList()); CombineInstance combineInstance = new CombineInstance(); combineInstance.transform = meshRenderer.transform.localToWorldMatrix; combineInstance.subMeshIndex = s; combineInstance.mesh = meshFilter.sharedMesh; (combineInstanceArrays[materialArrayIndex] as ArrayList).Add(combineInstance); }
我们看到上面有个函数Contains它就是用于判断是否有相同的材质,如果有就作为一张材质,Contains函数代码如下所示:
private int Contains(ArrayList searchList, string searchName) { for (int i = 0; i < searchList.Count; i++) { if (((Material)searchList[i]).name == searchName) { return i; } } return -1; }
作用是用于返回相同材质的索引,效果如下:
看上图红色的部分,有两个材质是一样的,材质只显示一个了,将其优化了。整个静态对象的优化就给大家解答完了。
时间: 2024-11-07 16:55:34