Net 4.6 RC x64的速度是x86(发行版)的两倍:
考虑这段代码:
class SpectralNorm
{
public static void Main(String[] args)
{
int n = 5500;
if (args.Length > 0) n = Int32.Parse(args[0]);
var spec = new SpectralNorm();
var watch = Stopwatch.StartNew();
var res = spec.Approximate(n);
Console.WriteLine("{0:f9} -- {1}", res, watch.Elapsed.TotalMilliseconds);
}
double Approximate(int n)
{
// create unit vector
double[] u = new double[n];
for (int i = 0; i < n; i++) u[i] = 1;
// 20 steps of the power method
double[] v = new double[n];
for (int i = 0; i < n; i++) v[i] = 0;
for (int i = 0; i < 10; i++)
{
MultiplyAtAv(n, u, v);
MultiplyAtAv(n, v, u);
}
// B=AtA A multiplied by A transposed
// v.Bv /(v.v) eigenvalue of v
double vBv = 0, vv = 0;
for (int i = 0; i < n; i++)
{
vBv += u[i] * v[i];
vv += v[i] * v[i];
}
return Math.Sqrt(vBv / vv);
}
/* return element i,j of infinite matrix A */
double A(int i, int j)
{
return 1.0 / ((i + j) * (i + j + 1) / 2 + i + 1);
}
/* multiply vector v by matrix A */
void MultiplyAv(int n, double[] v, double[] Av)
{
for (int i = 0; i < n; i++)
{
Av[i] = 0;
for (int j = 0; j < n; j++) Av[i] += A(i, j) * v[j];
}
}
/* multiply vector v by matrix A transposed */
void MultiplyAtv(int n, double[] v, double[] Atv)
{
for (int i = 0; i < n; i++)
{
Atv[i] = 0;
for (int j = 0; j < n; j++) Atv[i] += A(j, i) * v[j];
}
}
/* multiply vector v by matrix A and then by matrix A transposed */
void MultiplyAtAv(int n, double[] v, double[] AtAv)
{
double[] u = new double[n];
MultiplyAv(n, v, u);
MultiplyAtv(n, u, AtAv);
}
}
在我的机器上,x86发行版需要4.5秒才能完成,而x64则需要9.5秒。 x64是否需要任何特定的标志/设置?
更新
事实证明,RyuJIT在此问题中起作用。
如果在app.config中启用了
useLegacyJit
,结果将有所不同,这次x64的速度更快。<?xml version="1.0" encoding="utf-8"?>
<configuration>
<startup>
<supportedRuntime version="v4.0" sku=".NETFramework,Version=v4.6"/>
</startup>
<runtime>
<useLegacyJit enabled="1" />
</runtime>
</configuration>
更新
现在已将问题报告给CLR团队coreclr, issue 993
最佳答案
性能退化的原因在GitHub上得到了回答;简而言之,它似乎只能在Intel而不是Amd64机器上进行复制。内循环操作
Av[i] += v[j] * A(i, j);
结果是
IN002a: 000093 lea eax, [rax+r10+1]
IN002b: 000098 cvtsi2sd xmm1, rax
IN002c: 00009C movsd xmm2, qword ptr [@RWD00]
IN002d: 0000A4 divsd xmm2, xmm1
IN002e: 0000A8 movsxd eax, edi
IN002f: 0000AB movaps xmm1, xmm2
IN0030: 0000AE mulsd xmm1, qword ptr [r8+8*rax+16]
IN0031: 0000B5 addsd xmm0, xmm1
IN0032: 0000B9 movsd qword ptr [rbx], xmm0
Cvtsi2sd进行了低8字节的部分写入,而xmm寄存器的高字节未修改。对于repro情况,xmm1是部分编写的,但在代码中还有xmm1的进一步使用。这会在cvtsi2sd和其他使用xmm1的指令之间创建错误的依赖关系,从而影响指令的并行性。实际上,在cvtsi2sd修复了perf回归之前,将Int的代码源修改为Float强制转换以发出“xorps xmm1,xmm1”。
解决方法:如果我们在MultiplyAv/MultiplyAvt方法中的乘法运算中反转操作数的顺序,也可以避免Perf回归
void MultiplyAv(int n, double[] v, double[] Av)
{
for (int i = 0; i < n; i++)
{
Av[i] = 0;
for (int j = 0; j < n; j++)
Av[i] += v[j] * A(i, j); // order of operands reversed
}
}