与金属swift并行计算数组值的总和

Question

我试图与金属swift并行计算大数组的总和。

有神的方法吗？

我的平面是我将我的数组划分为子数组，并行计算一个子数组的总和，然后当并行计算完成时计算子和的总和。

例如，如果我有

array = [a0,....an] 

我在子数组中划分数组：

array_1 = [a_0,...a_i],
array_2 = [a_i+1,...a_2i],
....
array_n/i = [a_n-1, ... a_n]

这个数组的总和是并行计算的，我得到了

sum_1, sum_2, sum_3, ... sum_n/1

最后只计算子和的总和。

我创建运行我的金属着色器的应用程序，但有些事情我不太了解。

        var array:[[Float]] = [[1,2,3], [4,5,6], [7,8,9]]

        // get device
        let device: MTLDevice! = MTLCreateSystemDefaultDevice()

        // get library
        let defaultLibrary:MTLLibrary! = device.newDefaultLibrary()

        // queue
        let commandQueue:MTLCommandQueue! = device.newCommandQueue()

        // function
        let kernerFunction: MTLFunction! = defaultLibrary.newFunctionWithName("calculateSum")

        // pipeline with function
        let pipelineState: MTLComputePipelineState! = try device.newComputePipelineStateWithFunction(kernerFunction)

        // buffer for function
        let commandBuffer:MTLCommandBuffer! = commandQueue.commandBuffer()

        // encode function
        let commandEncoder:MTLComputeCommandEncoder = commandBuffer.computeCommandEncoder()

        // add function to encode
        commandEncoder.setComputePipelineState(pipelineState)

        // options
        let resourceOption = MTLResourceOptions()

        let arrayBiteLength = array.count * array[0].count * sizeofValue(array[0][0])

        let arrayBuffer = device.newBufferWithBytes(&array, length: arrayBiteLength, options: resourceOption)

        commandEncoder.setBuffer(arrayBuffer, offset: 0, atIndex: 0)

        var result:[Float] = [0,0,0]

        let resultBiteLenght = sizeofValue(result[0])

        let resultBuffer = device.newBufferWithBytes(&result, length: resultBiteLenght, options: resourceOption)

        commandEncoder.setBuffer(resultBuffer, offset: 0, atIndex: 1)

        let threadGroupSize = MTLSize(width: 1, height: 1, depth: 1)

        let threadGroups = MTLSize(width: (array.count), height: 1, depth: 1)

        commandEncoder.dispatchThreadgroups(threadGroups, threadsPerThreadgroup: threadGroupSize)

        commandEncoder.endEncoding()

        commandBuffer.commit()

        commandBuffer.waitUntilCompleted()

        let data = NSData(bytesNoCopy: resultBuffer.contents(), length: sizeof(Float), freeWhenDone: false)

        data.getBytes(&result, length: result.count * sizeof(Float))

        print(result)

是我的Swift代码，

我的着色器是：

kernel void calculateSum(const device float *inFloat [[buffer(0)]],
                     device float *result [[buffer(1)]],
                     uint id [[ thread_position_in_grid ]]) {


    float * f = inFloat[id];
    float sum = 0;
    for (int i = 0 ; i < 3 ; ++i) {
        sum = sum + f[i];
    }

    result = sum;
}

我不知道如何定义inFloat是数组数组。我不确切知道什么是threadGroupSize和threadGroups。我不知道着色器属性中的设备和uint是什么。

这是正确的方法吗？

Answer 1

我花时间用Metal创建了这个问题的完整工作示例。解释在评论中：

let count = 10_000_000
let elementsPerSum = 10_000

// Data type, has to be the same as in the shader
typealias DataType = CInt

let device = MTLCreateSystemDefaultDevice()!
let library = self.library(device: device)
let parsum = library.makeFunction(name: "parsum")!
let pipeline = try! device.makeComputePipelineState(function: parsum)

// Our data, randomly generated:
var data = (0..<count).map{ _ in DataType(arc4random_uniform(100)) }
var dataCount = CUnsignedInt(count)
var elementsPerSumC = CUnsignedInt(elementsPerSum)
// Number of individual results = count / elementsPerSum (rounded up):
let resultsCount = (count + elementsPerSum - 1) / elementsPerSum

// Our data in a buffer (copied):
let dataBuffer = device.makeBuffer(bytes: &data, length: MemoryLayout<DataType>.stride * count, options: [])!
// A buffer for individual results (zero initialized)
let resultsBuffer = device.makeBuffer(length: MemoryLayout<DataType>.stride * resultsCount, options: [])!
// Our results in convenient form to compute the actual result later:
let pointer = resultsBuffer.contents().bindMemory(to: DataType.self, capacity: resultsCount)
let results = UnsafeBufferPointer<DataType>(start: pointer, count: resultsCount)

let queue = device.makeCommandQueue()!
let cmds = queue.makeCommandBuffer()!
let encoder = cmds.makeComputeCommandEncoder()!

encoder.setComputePipelineState(pipeline)

encoder.setBuffer(dataBuffer, offset: 0, index: 0)

encoder.setBytes(&dataCount, length: MemoryLayout<CUnsignedInt>.size, index: 1)
encoder.setBuffer(resultsBuffer, offset: 0, index: 2)
encoder.setBytes(&elementsPerSumC, length: MemoryLayout<CUnsignedInt>.size, index: 3)

// We have to calculate the sum `resultCount` times => amount of threadgroups is `resultsCount` / `threadExecutionWidth` (rounded up) because each threadgroup will process `threadExecutionWidth` threads
let threadgroupsPerGrid = MTLSize(width: (resultsCount + pipeline.threadExecutionWidth - 1) / pipeline.threadExecutionWidth, height: 1, depth: 1)

// Here we set that each threadgroup should process `threadExecutionWidth` threads, the only important thing for performance is that this number is a multiple of `threadExecutionWidth` (here 1 times)
let threadsPerThreadgroup = MTLSize(width: pipeline.threadExecutionWidth, height: 1, depth: 1)

encoder.dispatchThreadgroups(threadgroupsPerGrid, threadsPerThreadgroup: threadsPerThreadgroup)
encoder.endEncoding()

var start, end : UInt64
var result : DataType = 0

start = mach_absolute_time()
cmds.commit()
cmds.waitUntilCompleted()
for elem in results {
    result += elem
}

end = mach_absolute_time()

print("Metal result: \(result), time: \(Double(end - start) / Double(NSEC_PER_SEC))")
result = 0

start = mach_absolute_time()
data.withUnsafeBufferPointer { buffer in
    for elem in buffer {
        result += elem
    }
}
end = mach_absolute_time()

print("CPU result: \(result), time: \(Double(end - start) / Double(NSEC_PER_SEC))")

我用我的Mac测试它，但它应该在iOS上运行得很好。

输出：

Metal result: 494936505, time: 0.024611456
CPU result: 494936505, time: 0.163341018

Metal版本的速度提高了约7倍。我敢肯定，如果你实施像截断或其他任何东西的分而治之，你可以获得更快的速度。

Answer 2

我一直在运行应用程序。在一个gt 740（384核）与i7-4790与多线程矢量和实现，这是我的数字：

Metal lap time: 19.959092
cpu MT lap time: 4.353881

这是cpu的5/1比例，所以除非你有一个强大的使用着色器的gpu是不值得的。

我一直在i7-3610qm w / igpu intel hd 4000上测试相同的代码，结果出乎意料的是金属效果更好：2/1

编辑：用线程参数调整后我终于提高了gpu性能，现在它高达16xcpu