c++ - 为什么 valarray 这么慢？

Question

我正在尝试使用 valarray，因为它在操作向量和矩阵时很像 MATLAB。我首先做了一些性能检查，发现 valarray 无法达到Stroustrup 在C++ 编程语言一书中声明的性能。

测试程序实际上做了 500 万次双倍乘法。我认为 c = a*b 至少可以与for循环双类型元素乘法相媲美，但我完全错了。我尝试了几台计算机和 Microsoft Visual C++ 6.0 和 Visual Studio 2008。

顺便说一句，我使用以下代码在 MATLAB 上进行了测试：

len = 5*1024*1024;
a = rand(len, 1);
b = rand(len, 1);
c = zeros(len, 1);
tic;
c = a.*b;
toc;

结果是 46 毫秒。这个时间精度不高；它仅作为参考。

代码是：

#include <iostream>
#include <valarray>
#include <iostream>
#include "windows.h"

using namespace std;
SYSTEMTIME stime;
LARGE_INTEGER sys_freq;

double gettime_hp();

int main()
{
    enum { N = 5*1024*1024 };
    valarray<double> a(N), b(N), c(N);
    QueryPerformanceFrequency(&sys_freq);
    int i, j;
    for (j=0 ; j<8 ; ++j)
    {
        for (i=0 ; i<N ; ++i)
        {
            a[i] = rand();
            b[i] = rand();
        }

        double* a1 = &a[0], *b1 = &b[0], *c1 = &c[0];
        double dtime = gettime_hp();
        for (i=0 ; i<N ; ++i)
            c1[i] = a1[i] * b1[i];
        dtime = gettime_hp()-dtime;
        cout << "double operator* " << dtime << " ms\n";

        dtime = gettime_hp();
        c = a*b ;
        dtime = gettime_hp() - dtime;
        cout << "valarray operator* " << dtime << " ms\n";

        dtime = gettime_hp();
        for (i=0 ; i<N ; ++i)
            c[i] = a[i] * b[i];
        dtime = gettime_hp() - dtime;
        cout << "valarray[i] operator* " << dtime<< " ms\n";

        cout << "------------------------------------------------------\n";
    }
}

double gettime_hp()
{
    LARGE_INTEGER tick;
    extern LARGE_INTEGER sys_freq;
    QueryPerformanceCounter(&tick);
    return (double)tick.QuadPart * 1000.0 / sys_freq.QuadPart;
}

运行结果：（最大速度优化的发布模式）

double operator* 52.3019 ms
valarray operator* 128.338 ms
valarray[i] operator* 43.1801 ms
------------------------------------------------------
double operator* 43.4036 ms
valarray operator* 145.533 ms
valarray[i] operator* 44.9121 ms
------------------------------------------------------
double operator* 43.2619 ms
valarray operator* 158.681 ms
valarray[i] operator* 43.4871 ms
------------------------------------------------------
double operator* 42.7317 ms
valarray operator* 173.164 ms
valarray[i] operator* 80.1004 ms
------------------------------------------------------
double operator* 43.2236 ms
valarray operator* 158.004 ms
valarray[i] operator* 44.3813 ms
------------------------------------------------------

具有相同优化的调试模式：

double operator* 41.8123 ms
valarray operator* 201.484 ms
valarray[i] operator* 41.5452 ms
------------------------------------------------------
double operator* 40.2238 ms
valarray operator* 215.351 ms
valarray[i] operator* 40.2076 ms
------------------------------------------------------
double operator* 40.5859 ms
valarray operator* 232.007 ms
valarray[i] operator* 40.8803 ms
------------------------------------------------------
double operator* 40.9734 ms
valarray operator* 234.325 ms
valarray[i] operator* 40.9711 ms
------------------------------------------------------
double operator* 41.1977 ms
valarray operator* 234.409 ms
valarray[i] operator* 41.1429 ms
------------------------------------------------------
double operator* 39.7754 ms
valarray operator* 234.26 ms
valarray[i] operator* 39.6338 ms
------------------------------------------------------

score 24 · Accepted Answer

我刚刚在 Linux x86-64 系统（Sandy Bridge CPU）上试过：

gcc 4.5.0：

double operator* 9.64185 ms
valarray operator* 9.36987 ms
valarray[i] operator* 9.35815 ms

英特尔 ICC 12.0.2：

double operator* 7.76757 ms
valarray operator* 9.60208 ms
valarray[i] operator* 7.51409 ms

在这两种情况下，我只使用了-O3其他与优化相关的标志。

看起来 MS C++ 编译器和/或 valarray 实现很糟糕。

这是针对 Linux 修改的 OP 代码：

#include <iostream>
#include <valarray>
#include <iostream>
#include <ctime>

using namespace std ;

double gettime_hp();

int main()
{
    enum { N = 5*1024*1024 };
    valarray<double> a(N), b(N), c(N) ;
    int i,j;
    for(  j=0 ; j<8 ; ++j )
    {
        for(  i=0 ; i<N ; ++i )
        {
            a[i]=rand();
            b[i]=rand();
        }

        double* a1 = &a[0], *b1 = &b[0], *c1 = &c[0] ;
        double dtime=gettime_hp();
        for(  i=0 ; i<N ; ++i ) c1[i] = a1[i] * b1[i] ;
        dtime=gettime_hp()-dtime;
        cout << "double operator* " << dtime << " ms\n" ;

        dtime=gettime_hp();
        c = a*b ;
        dtime=gettime_hp()-dtime;
        cout << "valarray operator* " << dtime << " ms\n" ;

        dtime=gettime_hp();
        for(  i=0 ; i<N ; ++i ) c[i] = a[i] * b[i] ;
        dtime=gettime_hp()-dtime;
        cout << "valarray[i] operator* " << dtime<< " ms\n" ;

        cout << "------------------------------------------------------\n" ;
    }
}

double gettime_hp()
{
    struct timespec timestamp;

    clock_gettime(CLOCK_REALTIME, &timestamp);
    return timestamp.tv_sec * 1000.0 + timestamp.tv_nsec * 1.0e-6;
}

score 13 · Accepted Answer

我怀疑原因c = a*b比一次执行一个元素的操作慢得多是因为

template<class T> valarray<T> operator*
    (const valarray<T>&, const valarray<T>&);

运算符必须分配内存以将结果放入，然后按值返回。

即使使用“交换”来执行复制，该功能仍然具有

为结果分配新块valarray
初始化新valarray的（这可能会被优化掉）
将结果放入新的valarray
在内存中为新的内存分页，valarray因为它被初始化或使用结果值设置
valarray释放被结果替换的旧的

score 6 · Accepted Answer

valarray 的重点是在向量机上快速，而 x86 机器却不是。

非向量机上的一个好的实现应该能够与您获得的性能相匹配，例如

for (i=0; i < N; ++i) 
    c1[i] = a1[i] * b1[i];

一个坏的当然不会。除非硬件中有一些东西可以加速并行处理，否则这将非常接近你能做到的最好。

score 3 · Accepted Answer

我终于通过使用延迟评估得到了这个。代码可能很难看，因为我刚刚开始学习这些 C++ 高级概念。

这是代码：

#include <iostream>
#include <valarray>
#include <iostream>
#include "windows.h"

using namespace std;
SYSTEMTIME stime;
LARGE_INTEGER sys_freq;

double gettime_hp();

// To improve the c = a*b (it will generate a temporary first, assigned to 'c' and delete the temporary.
// Which causes the program really slow
// The solution is the expression template and let the compiler to decide when all the expression is known.


// Delayed evaluation
//typedef valarray<double> Vector;
class Vector;

class VecMul
{
    public:
        const Vector& va;
        const Vector& vb;
        //Vector& vc;
        VecMul(const Vector& v1, const Vector& v2): va(v1), vb(v2) {}
        operator Vector();
};

class Vector:public valarray<double>
{
    valarray<double> *p;

    public:
        explicit Vector(int n)
        {
            p = new valarray<double>(n);
        }
        Vector& operator = (const VecMul &m)
        {
            for(int i=0; i<m.va.size(); i++)
                (*p)[i] = (m.va)[i]*(m.vb)[i]; // Ambiguous
            return *this;
        }
        double& operator[](int i) const {return (*p)[i];} //const vector_type[i]
        int size()const {return (*p).size();}
};


inline VecMul operator*(const Vector& v1, const Vector& v2)
{
    return VecMul(v1, v2);
}


int main()
{
    enum {N = 5*1024*1024};
    Vector a(N), b(N), c(N);
    QueryPerformanceFrequency(&sys_freq);
    int i, j;
    for (j=0 ; j<8 ; ++j)
    {
        for (i=0 ; i<N ; ++i)
        {
            a[i] = rand();
            b[i] = rand();
        }

        double* a1 = &a[0], *b1 = &b[0], *c1 = &c[0];
        double dtime = gettime_hp();
        for (i=0 ; i<N ; ++i)
            c1[i] = a1[i] * b1[i];
        dtime = gettime_hp()-dtime;
        cout << "double operator* " << dtime << " ms\n";

        dtime = gettime_hp();
        c = a*b;
        dtime = gettime_hp()-dtime;
        cout << "valarray operator* " << dtime << " ms\n";

        dtime = gettime_hp();
        for (i=0 ; i<N ; ++i)
            c[i] = a[i] * b[i];
        dtime = gettime_hp() - dtime;
        cout << "valarray[i] operator* " << dtime << " ms\n";

        cout << "------------------------------------------------------\n";
    }
}

double gettime_hp()
{
    LARGE_INTEGER tick;
    extern LARGE_INTEGER sys_freq;
    QueryPerformanceCounter(&tick);
    return (double)tick.QuadPart*1000.0/sys_freq.QuadPart;
}

在 Visual Studio 上的运行结果是：

double operator* 41.2031 ms
valarray operator* 43.8407 ms
valarray[i] operator* 42.49 ms

score 1 · Accepted Answer

我在 Visual Studio 2010 x64 版本中进行编译。我对您的代码进行了非常轻微的更改：

    double* a1 = &a[0], *b1 = &b[0], *c1 = &c[0];
    double dtime = gettime_hp();
    for (i=0 ; i<N ; ++i)
        a1[i] *= b1[i];
    dtime = gettime_hp() - dtime;
    cout << "double operator* " << dtime << " ms\n";

    dtime = gettime_hp();
    a *= b;
    dtime = gettime_hp() - dtime;
    cout << "valarray operator* " << dtime << " ms\n";

    dtime = gettime_hp();
    for (i=0 ; i<N ; ++i)
        a[i] *= b[i];
    dtime = gettime_hp() - dtime;
    cout << "valarray[i] operator* " << dtime<< " ms\n";

    cout << "------------------------------------------------------\n" ;

在这里你可以看到我使用 *= 而不是c = a * b. 在更现代的数学库中，使用了非常复杂的表达式模板机制来消除这个问题。在这种情况下，我实际上从 valarray 获得了稍微快一点的结果，尽管这可能只是因为内容已经在缓存中。您看到的开销只是多余的临时变量，而 valarray 没有任何内在的东西，特别是 - 您会看到类似std::string.

score -1 · Accepted Answer

我认为 Michael Burr 的回答是正确的。也许您可以创建一个虚拟类型作为 operator 的返回值的类型，然后为这个虚拟类型+重新加载另一个类型（粗略地说）。operator=operator=(virtual type& v){&valarray=&v;v=NULL;}

当然，在 valarray 上实现这个想法是很困难的。但是当你创建一个新类时，你可以试试这个想法。然后，效率operator+几乎与相同operator+=。

score -2 · Accepted Answer

嗯..我测试了Blitz++，它和 valarray 一样......而且，Blitz++[]运算符非常慢。

#include <blitz/array.h>
#include <iostream>

#ifdef WIN32
#include "windows.h"
LARGE_INTEGER sys_freq;
#endif

#ifdef LINUX
<ctime>
#endif

using namespace std;
SYSTEMTIME stime;

__forceinline double gettime_hp();
double gettime_hp()
{
    #ifdef WIN32
        LARGE_INTEGER tick;
        extern LARGE_INTEGER sys_freq;
        QueryPerformanceCounter(&tick);
        return (double)tick.QuadPart * 1000.0 / sys_freq.QuadPart;
    #endif

    #ifdef LINUX
        struct timespec timestamp;

        clock_gettime(CLOCK_REALTIME, &timestamp);
        return timestamp.tv_sec * 1000.0 + timestamp.tv_nsec * 1.0e-6;
    #endif
}
BZ_USING_NAMESPACE(blitz)

int main()
{
    int N = 5*1024*1024;

    // Create three-dimensional arrays of double
    Array<double, 1> a(N), b(N), c(N);

    int i, j;

    #ifdef WIN32
        QueryPerformanceFrequency(&sys_freq);
    #endif

    for (j=0 ; j<8 ; ++j)
    {
        for (i=0 ; i<N ; ++i)
        {
            a[i] = rand();
            b[i] = rand();
        }

        double* a1 = a.data(), *b1 = b.data(), *c1 = c.data();
        double dtime = gettime_hp();
        for (i=0 ; i<N ; ++i)
            c1[i] = a1[i] * b1[i];
        dtime = gettime_hp() - dtime;
        cout << "double operator* " << dtime << " ms\n";

        dtime = gettime_hp();
        c = a*b;
        dtime = gettime_hp() - dtime;
        cout << "blitz operator* " << dtime << " ms\n";

        dtime = gettime_hp();
        for (i=0 ; i<N ; ++i)
            c[i] = a[i] * b[i];
        dtime = gettime_hp() - dtime;
        cout << "blitz[i] operator* " << dtime<< " ms\n";

        cout << "------------------------------------------------------\n";
    }
}

c++ - 为什么 valarray 这么慢？

7 回答 7

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