Question

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需要9.5秒。 x64是否需要特定的标志/设置？

更新

事实证明，RyuJIT在这个问题上发挥了作用。如果在app.config中启用了useLegacyJit，则结果会有所不同，这次x64会更快。

<?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

现在已向CLR小组coreclr, issue 993

报告此问题

Answer 1

在GitHub上回答了完成回归的原因;简而言之，它似乎只在英特尔而不是在Amd64机器上重现。内循环操作

Av[i] += v[j] * A(i, j);

结果

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对较低的8字节进行部分写入，xmm寄存器的高位字节未经修改。对于repro情况，xmm1是部分写入的，但是代码中还有xmm1的进一步用途。这会在cvtsi2sd和使用xmm1的其他指令之间产生错误的依赖关系，这会影响指令的并行性。确实修改了Int to Float的codegen以在cvtsi2sd修复perf回归之前发出“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
    }
}

.NET 4.6 RC x64的速度是x86（发行版）

1 个答案: