与Metal Swift并行计算数组值的总和 [英] Compute sum of array values in parallel with metal swift
问题描述
我正在尝试与金属swift并行计算大型数组的总和.
I am trying to compute sum of large array in parallel with metal swift.
有神的方法吗?
我的平面是我将数组划分为子数组,并行计算一个子数组的和,然后在并行计算完成时计算子和的和.
My plane was that I divide my array to sub arrays, compute sum of one sub arrays in parallel and then when parallel computation is finished compute sum of sub sums.
例如,如果我有
array = [a0,....an]
我将数组划分为子数组:
I divide array in sub arrays :
array_1 = [a_0,...a_i],
array_2 = [a_i+1,...a_2i],
....
array_n/i = [a_n-1, ... a_n]
此数组的总和是并行计算的,我得到
sums for this arrays is computed in parallel and I get
sum_1, sum_2, sum_3, ... sum_n/1
最后只计算子和.
我创建了运行金属着色器的应用程序,但是有些事情我不太了解.
I create application which run my metal shader, but some things I don't understand quite.
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,
is my Swift code,
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;
}
我不知道如何将inFloat定义为array的数组. 我不确切地知道什么是threadGroupSize和threadGroups. 我不知道着色器属性中的设备和uint是什么.
I don't know how to defined that inFloat is array of array. I don't know exactly what is threadGroupSize and threadGroups. I don't know what is device and uint in shader properties.
这是正确的方法吗?
推荐答案
我花了一些时间来创建有关Metal的此问题的完整示例.注释中有解释:
I took the time to create a fully working example of this problem with Metal. The explanation is in the comments:
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))")
我使用Mac进行了测试,但它在iOS上应该可以正常工作.
I used my Mac to test it, but it should work just fine on iOS.
输出:
Metal result: 494936505, time: 0.024611456
CPU result: 494936505, time: 0.163341018
金属版本的速度提高了约7倍.我敢肯定,如果您使用截止或诸如此类的方法实现分而治之,那么您可以提高速度.
The Metal version is about 7 times faster. I'm sure you can get more speed if you implement something like divide-and-conquer with cutoff or whatever.
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