小编Rub*_*xis的帖子

这是我在OpenGL中使用的两种混合模式,即IOS中对金属的转换

glEnable(GL_BLEND);

glBlendFuncSeparate(GL_SRC_ALPHA,GL_ONE_MINUS_SRC_ALPHA,GL_ONE,GL_ONE_MINUS_SRC_ALPHA);

glBlendFuncSeparate(GL_SRC_ALPHA, GL_ONE, GL_SRC_ALPHA, GL_ONE);

 func mtkView(_ view: MTKView, drawableSizeWillChange size: CGSize) {
        print("current drawable size:\(view.drawableSize)")
    }

    func draw(in view: MTKView) {


        guard let drawable = view.currentDrawable else { return }

        let textureDescriptor = MTLTextureDescriptor()
        textureDescriptor.textureType = MTLTextureType.type2D
        textureDescriptor.width = drawable.texture.width
        textureDescriptor.height = drawable.texture.height
        textureDescriptor.pixelFormat = .bgra8Unorm
        textureDescriptor.storageMode = .shared
        textureDescriptor.usage = .renderTarget

        let sampleTexture = device.makeTexture(descriptor: textureDescriptor)


        let renderPass = MTLRenderPassDescriptor()
        renderPass.colorAttachments[0].texture = sampleTexture
        renderPass.colorAttachments[0].loadAction = .clear
        renderPass.colorAttachments[0].clearColor =
            MTLClearColor(red: 1.0, green: 0.0, blue: 0.0, alpha: 1.0)
        renderPass.colorAttachments[0].storeAction = .store

        let commandBuffer = commandQueue.makeCommandBuffer() …

我正在从 obj 文件绘制一个 3D 模型，如下所示。我需要找到该 obj 文件的每个子网格物体的边界框。

 let assetURL = Bundle.main.url(forResource: "train", withExtension: "obj")!
 let allocator = MTKMeshBufferAllocator(device: device)
 let asset = MDLAsset(url: assetURL, vertexDescriptor: vertexDescriptor, bufferAllocator: meshAllocator)
 let mdlMesh = asset.object(at: 0) as! MDLMesh

 mesh = try! MTKMesh(mesh: mdlMesh, device: device)

  for submesh in mesh.submeshes {
//I need to find the bounding box for each Submesh
        }

我怎样才能在 iOS 中实现这一点。

使用 Metal 我使用四个点使用贝塞尔曲线绘制线。我为线条使用了近 1500 个三角形。该线是像素化的。我怎样才能减少像素化。

vertex VertexOutBezier bezier_vertex(constant BezierParameters *allParams[[buffer(0)]],
                               constant GlobalParameters& globalParams[[buffer(1)]],
                               uint vertexId [[vertex_id]],
                               uint instanceId [[instance_id]])
{

    float t = (float) vertexId / globalParams.elementsPerInstance;


    rint(t);
    BezierParameters params = allParams[instanceId];

    float lineWidth = (1 - (((float) (vertexId % 2)) * 2.0)) * params.lineThickness;
 float2 a = params.a;
    float2 b = params.b;

    float cx = distance(a , b);


float2 p1 = params.p1 * 3.0;  // float2 p1 = params.p1 * 3.0;
    float2 p2 = params.p2 * 3.0;  // float2 …

在Obj CI中创建了一个属性

@property(nonatomic) UILabel *subscriptionText;

然后我为该属性UILabel创建了一个setter方法.如下

-(UILabel *)subscriptionText{
    if (!_subscriptionText) {
        _subscriptionText = [UILabel new];
        _subscriptionText.translatesAutoresizingMaskIntoConstraints = NO;
   _subscriptionText.textAlignment = NSTextAlignmentJustified;


    }
    return _subscriptionText;
}

然后在viewdidload中我添加此视图

[self.view addSubview:self.subscriptionText];

如何在Swift 4.2中执行相同的操作.

我是金属新手。我们可以在单个应用程序中创建多少个 MTLRenderpassdescriptor（任意限制）？

我想要三个 MTLRenderpassdescriptor 是否可能。一我需要根据手指位置绘制四边形。这就像离屏渲染。将会有纹理输出。

我想将输出纹理传递给另一个渲染通道描述符来处理它的颜色，这也类似于离屏渲染。这里将提供新的输出纹理

我想向屏幕呈现新的输出纹理。是否可以 ？当我创建三个渲染通道描述符时，只有两个可用。

第一次离屏渲染效果很好

let renderPass = MTLRenderPassDescriptor()

renderPass.colorAttachments[0].texture = ssTexture
renderPass.colorAttachments[0].loadAction = .clear
renderPass.colorAttachments[0].clearColor = MTLClearColorMake(  0.0,  0.0,  0.0,  0.0)
renderPass.colorAttachments[0].storeAction = .store
let commandBuffer = commandQueue.makeCommandBuffer()

var commandEncoder = commandBuffer?.makeRenderCommandEncoder(descriptor: renderPass)
for scene in scenesTool{
    scene.render(commandEncoder: commandEncoder!)
}
commandEncoder?.endEncoding()

此渲染通道描述符在调试中也不可用。

let renderPassWC = MTLRenderPassDescriptor()
renderPassWC.colorAttachments[0].texture = ssCurrentTexture
renderPassWC.colorAttachments[0].loadAction = .clear
renderPassWC.colorAttachments[0].clearColor = MTLClearColorMake(  0.0,  0.0,  0.0,  0.0)
renderPassWC.colorAttachments[0].storeAction = .store

let commandBufferWC = commandQueue.makeCommandBuffer()
let commandEncoderWC = commandBufferWC?.makeRenderCommandEncoder(descriptor: renderPassWC)
for scene in scenesWCTool{ …

fragment half4 fragmen_shader_test(WaterColorCloudOut params[[stage_in]],
                                                         texture2d<float , access::sample>cloud1 [[texture(0)]],
                                                         texture2d<half, access::sample> cloud2 [[texture(1)]],
                                                         texture2d<half, access::sample> cloud3 [[texture(2)]]
                                                         )
{
    constexpr sampler defaultSampler;
    float4 color1;

    if(params.index == 0){
        color1= float4(cloud1.sample(defaultSampler, float2(params.textureCoordinates)))  * params.color ;
    }
    else if(params.index == 1){
        color1= float4(cloud2.sample(defaultSampler, float2(params.textureCoordinates)))  * params.color ;
    } else{
        color1= float4(cloud3.sample(defaultSampler, float2(params.textureCoordinates)))  * params.color ;
    }

 return half4(color1);
}

这里我使用了三个纹理，因为 If-else 条件的性能随着时间的推移而下降。我觉得如果我将纹理数组发送到着色器，则无需执行 if else 语句。在CPU中我有三个MTLTexture。如何将三个纹理绑定到一个数组并将其传递给着色器。

在CPU端我创建了三个纹理并创建了一个MTLTexture数组

 var textureArray:[MTLTexture] = [] 

然后我将纹理附加到该数组。在 MTLRenderCommandEncoder 中

let myRange: CountableRange = 0..<2
commandEncoder.setFragmentTextures(textureArray, range: myRange)

在着色器中

texture2d_array<float , …