c#-矢量弱SIMD性能
作者:互联网
我正在优化算法,正在考虑将Vector应用于double进行乘法和累加运算.最接近的实现显然是Vector.dot(v1,v2); …但是,为什么我的代码这么慢?
namespace ConsoleApp1 {
class Program {
public static double SIMDMultAccumulate(double[] inp1, double[] inp2) {
var simdLength = Vector<double>.Count;
var returnDouble = 0d;
// Find the max and min for each of Vector<ushort>.Count sub-arrays
var i = 0;
for (; i <= inp1.Length - simdLength; i += simdLength) {
var va = new Vector<double>(inp1, i);
var vb = new Vector<double>(inp2, i);
returnDouble += Vector.Dot(va, vb);
}
// Process any remaining elements
for (; i < inp1.Length; ++i) {
var va = new Vector<double>(inp1, i);
var vb = new Vector<double>(inp2, i);
returnDouble += Vector.Dot(va, vb);
}
return returnDouble;
}
public static double NonSIMDMultAccumulate(double[] inp1, double[] inp2) {
var returnDouble = 0d;
for (int i = 0; i < inp1.Length; i++) {
returnDouble += inp1[i] * inp2[i];
}
return returnDouble;
}
static void Main(string[] args) {
Console.WriteLine("Is hardware accelerated: " + Vector.IsHardwareAccelerated);
const int size = 24;
var inp1 = new double[size];
var inp2 = new double[size];
var random = new Random();
for (var i = 0; i < inp1.Length; i++) {
inp1[i] = random.NextDouble();
inp2[i] = random.NextDouble();
}
var sumSafe = 0d;
var sumFast = 0d;
var sw = Stopwatch.StartNew();
for (var i = 0; i < 10; i++) {
sumSafe = NonSIMDMultAccumulate(inp1, inp2);
}
Console.WriteLine("{0} Ticks", sw.Elapsed.Ticks);
sw.Restart();
for (var i = 0; i < 10; i++) {
sumFast = SIMDMultAccumulate(inp1, inp2);
}
Console.WriteLine("{0} Ticks", sw.Elapsed.Ticks);
// Assert.AreEqual(sumSafe, sumFast, 0.00000001);
}
}
}
与非SIMD版本相比,SIMD版本需要大约70%的刻度.我正在运行Haswell架构和恕我直言. FMA3应该被实施! (发布版本,首选x64).
有任何想法吗?
多谢你们!
解决方法:
使用BechmarkDotNet,假定输入数组的长度(ITEMS = 10000)是Vector.Count的倍数,使用SIMD Vector可以获得几乎两倍的性能:
[Benchmark(Baseline = true)]
public double DotDouble()
{
double returnVal = 0.0;
for(int i = 0; i < ITEMS; i++)
{
returnVal += doubleArray[i] * doubleArray2[i];
}
return returnVal;
}
[Benchmark]
public double DotDoubleVectorNaive()
{
double returnVal = 0.0;
for(int i = 0; i < ITEMS; i += doubleSlots)
{
returnVal += Vector.Dot(new Vector<double>(doubleArray, i), new Vector<double>(doubleArray2, i));
}
return returnVal;
}
[Benchmark]
public double DotDoubleVectorBetter()
{
Vector<double> sumVect = Vector<double>.Zero;
for (int i = 0; i < ITEMS; i += doubleSlots)
{
sumVect += new Vector<double>(doubleArray, i) * new Vector<double>(doubleArray2, i);
}
return Vector.Dot(sumVect, Vector<double>.One);
}
BenchmarkDotNet=v0.10.14, OS=Windows 10.0.17134
Intel Core i7-4500U CPU 1.80GHz (Haswell), 1 CPU, 4 logical and 2 physical cores
Frequency=1753758 Hz, Resolution=570.2041 ns, Timer=TSC
.NET Core SDK=2.1.300
[Host] : .NET Core 2.1.0 (CoreCLR 4.6.26515.07, CoreFX 4.6.26515.06), 64bit RyuJIT
DefaultJob : .NET Core 2.1.0 (CoreCLR 4.6.26515.07, CoreFX 4.6.26515.06), 64bit RyuJIT
Method | Mean | Error | StdDev | Scaled |
---------------------- |----------:|----------:|----------:|-------:|
DotDouble | 10.341 us | 0.0902 us | 0.0844 us | 1.00 |
DotDoubleVectorNaive | 5.907 us | 0.0206 us | 0.0183 us | 0.57 |
DotDoubleVectorBetter | 4.825 us | 0.0197 us | 0.0184 us | 0.47 |
为了完整起见,RiuJIT将在Haswell上将Vector.Dot产品编译为:
vmulpd ymm0,ymm0,ymm1
vhaddpd ymm0,ymm0,ymm0
vextractf128 xmm2,ymm0,1
vaddpd xmm0,xmm0,xmm2
vaddsd xmm6,xmm6,xmm0
编辑以根据注释添加点积在循环外的大小写,并为点积添加ASm.
标签:simd,c,system-numerics 来源: https://codeday.me/bug/20191108/2010051.html