金属中的大块渲染-Java 学习之路

我正在尝试使用Metal创建一个程序游戏，我正在使用基于八叉树的块方法来实现Level of Detail .

我正在使用的方法涉及CPU为地形创建八叉树节点，然后使用计算着色器在GPU上创建其网格 . 此网格存储在块对象中的顶点缓冲区和索引缓冲区中以进行渲染 .

所有这些看起来都运行得相当不错，但是在渲染块时我很早就遇到了性能问题 . 目前我收集了一个数组来绘制，然后将其提交给我的渲染器，这将创建一个 MTLParallelRenderCommandEncoder 然后为每个块创建一个 MTLRenderCommandEncoder ，然后将其提交给GPU .

从它的外观来看，大约50％的CPU时间用于为每个块创建 MTLRenderCommandEncoder . 目前我在这些早期阶段下降到50fps左右 . （实际上似乎每个 MTLParallelRenderCommandEncoder 中最多只能有63个 MTLRenderCommandEncoder 所以它不完全是4x4x4）

我已经读过 MTLParallelRenderCommandEncoder 的意思是在一个单独的线程中创建每个 MTLRenderCommandEncoder ，但是我会在63个块的上限周围渲染为最大值 .

我觉得以某种方式将每个块的顶点和索引缓冲区合并为一个或两个更大的缓冲区以提交将有所帮助，但我不知道如何在没有大量 memcpy() 调用的情况下执行此操作，以及这是否甚至可以提高效率 .

这是我的代码，它接收节点数组并绘制它们：

func drawNodes(nodes: [OctreeNode], inView view: AHMetalView){
  // For control of several rotating buffers
  dispatch_semaphore_wait(displaySemaphore, DISPATCH_TIME_FOREVER)

  makeDepthTexture()

  updateUniformsForView(view, duration: view.frameDuration)
  let commandBuffer = commandQueue.commandBuffer()


  let optDrawable = layer.nextDrawable()

  guard let drawable = optDrawable else{
    return
  }

  let passDescriptor = MTLRenderPassDescriptor()

  passDescriptor.colorAttachments[0].texture = drawable.texture
  passDescriptor.colorAttachments[0].clearColor = MTLClearColorMake(0.2, 0.2, 0.2, 1)
  passDescriptor.colorAttachments[0].storeAction = .Store
  passDescriptor.colorAttachments[0].loadAction = .Clear

  passDescriptor.depthAttachment.texture = depthTexture
  passDescriptor.depthAttachment.clearDepth = 1
  passDescriptor.depthAttachment.loadAction = .Clear
  passDescriptor.depthAttachment.storeAction = .Store

  let parallelRenderPass = commandBuffer.parallelRenderCommandEncoderWithDescriptor(passDescriptor)

  // Currently 63 nodes as a maximum
  for node in nodes{
    // This line is taking up around 50% of the CPU time
    let renderPass = parallelRenderPass.renderCommandEncoder()

    renderPass.setRenderPipelineState(renderPipelineState)
    renderPass.setDepthStencilState(depthStencilState)
    renderPass.setFrontFacingWinding(.CounterClockwise)
    renderPass.setCullMode(.Back)

    let uniformBufferOffset = sizeof(AHUniforms) * uniformBufferIndex

    renderPass.setVertexBuffer(node.vertexBuffer, offset: 0, atIndex: 0)
    renderPass.setVertexBuffer(uniformBuffer, offset: uniformBufferOffset, atIndex: 1)

    renderPass.setTriangleFillMode(.Lines)

    renderPass.drawIndexedPrimitives(.Triangle, indexCount: AHMaxIndicesPerChunk, indexType: AHIndexType, indexBuffer: node.indexBuffer, indexBufferOffset: 0)

    renderPass.endEncoding()
  }
  parallelRenderPass.endEncoding()

  commandBuffer.presentDrawable(drawable)

  commandBuffer.addCompletedHandler { (commandBuffer) -> Void in
    self.uniformBufferIndex = (self.uniformBufferIndex + 1) % AHInFlightBufferCount
    dispatch_semaphore_signal(self.displaySemaphore)
  }

  commandBuffer.commit()
}

1 回答

你注意到：

我已经读过MTLParallelRenderCommandEncoder的意思是在一个单独的线程中创建每个MTLRenderCommandEncoder ......

而你正在做的是顺序创建，编码和结束命令编码器 - 这里没有任何并行，所以 MTLParallelRenderCommandEncoder 对你没有任何作用 . 如果你消除了并行编码器并且在每次通过for循环时只用 renderCommandEncoderWithDescriptor(_:) 创建了编码器，那么你的性能大致相同......也就是说，由于创建的开销，你仍然会遇到相同的性能问题所有这些编码器 .

因此，如果您要按顺序编码，只需重复使用相同的编码器即可 . 此外，您应该尽可能多地重用其他共享状态 . 以下是可能的重构（未经测试）的快速通过：

let passDescriptor = MTLRenderPassDescriptor()

// call this once before your render loop
func setup() {
    makeDepthTexture()

    passDescriptor.colorAttachments[0].clearColor = MTLClearColorMake(0.2, 0.2, 0.2, 1)
    passDescriptor.colorAttachments[0].storeAction = .Store
    passDescriptor.colorAttachments[0].loadAction = .Clear

    passDescriptor.depthAttachment.texture = depthTexture
    passDescriptor.depthAttachment.clearDepth = 1
    passDescriptor.depthAttachment.loadAction = .Clear
    passDescriptor.depthAttachment.storeAction = .Store

    // set up render pipeline state and depthStencil state
}

func drawNodes(nodes: [OctreeNode], inView view: AHMetalView) {

    updateUniformsForView(view, duration: view.frameDuration)

    // Set up completed handler ahead of time
    let commandBuffer = commandQueue.commandBuffer()
    commandBuffer.addCompletedHandler { _ in // unused parameter
        self.uniformBufferIndex = (self.uniformBufferIndex + 1) % AHInFlightBufferCount
        dispatch_semaphore_signal(self.displaySemaphore)
    }

    // Semaphore should be tied to drawable acquisition
    dispatch_semaphore_wait(displaySemaphore, DISPATCH_TIME_FOREVER)
    guard let drawable = layer.nextDrawable()
        else { return }

    // Set up the one part of the pass descriptor that changes per-frame
    passDescriptor.colorAttachments[0].texture = drawable.texture

    // Get one render pass descriptor and reuse it
    let renderPass = commandBuffer.renderCommandEncoderWithDescriptor(passDescriptor)
    renderPass.setTriangleFillMode(.Lines)
    renderPass.setRenderPipelineState(renderPipelineState)
    renderPass.setDepthStencilState(depthStencilState)

    for node in nodes {
        // Update offsets and draw
        let uniformBufferOffset = sizeof(AHUniforms) * uniformBufferIndex
        renderPass.setVertexBuffer(node.vertexBuffer, offset: 0, atIndex: 0)
        renderPass.setVertexBuffer(uniformBuffer, offset: uniformBufferOffset, atIndex: 1)
        renderPass.drawIndexedPrimitives(.Triangle, indexCount: AHMaxIndicesPerChunk, indexType: AHIndexType, indexBuffer: node.indexBuffer, indexBufferOffset: 0)

    }
    renderPass.endEncoding()

    commandBuffer.presentDrawable(drawable)
    commandBuffer.commit()
}

然后，使用Instruments查看您可能遇到的进一步性能问题（如果有） . 有一个很好的WWDC 2015 session关于显示几个常见的"gotchas"，如何在分析中诊断它们，以及如何解决它们 .

回复于 2024-05-08T06:38:21+08:00

金属中的大块渲染

1 回答

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