如何实现每个周期4个FLOP的理论最大值？

Question

如何在现代x86-64 Intel CPU上实现每个周期4个浮点运算(双精度)的理论峰值性能？

据我所知,SSE 需要三个周期,add而mul大多数现代Intel CPU需要五个周期才能完成(参见例如Agner Fog的"指令表").由于流水线操作,add如果算法具有至少三个独立的求和,则每个周期可以获得一个吞吐量.因为打包addpd和标量addsd版本都是如此,并且SSE寄存器可以包含两个,double每个周期的吞吐量可以高达两个触发器.

此外,似乎(虽然我没有看到任何适当的文档)add并且mul可以并行执行,给出每个周期四个触发器的理论最大吞吐量.

但是,我无法使用简单的C/C++程序复制该性能.我最好的尝试导致大约2.7个翻牌/周期.如果有人可以贡献一个简单的C/C++或汇编程序,它可以表现出非常高兴的峰值性能.

我的尝试:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <sys/time.h>

double stoptime(void) {
   struct timeval t;
   gettimeofday(&t,NULL);
   return (double) t.tv_sec + t.tv_usec/1000000.0;
}

double addmul(double add, double mul, int ops){
   // Need to initialise differently otherwise compiler might optimise away
   double sum1=0.1, sum2=-0.1, sum3=0.2, sum4=-0.2, sum5=0.0;
   double mul1=1.0, mul2= 1.1, mul3=1.2, mul4= 1.3, mul5=1.4;
   int loops=ops/10;          // We have 10 floating point operations inside the loop
   double expected = 5.0*add*loops + (sum1+sum2+sum3+sum4+sum5)
               + pow(mul,loops)*(mul1+mul2+mul3+mul4+mul5);

   for (int i=0; i<loops; i++) {
      mul1*=mul; mul2*=mul; mul3*=mul; mul4*=mul; mul5*=mul;
      sum1+=add; sum2+=add; sum3+=add; sum4+=add; sum5+=add;
   }
   return  sum1+sum2+sum3+sum4+sum5+mul1+mul2+mul3+mul4+mul5 - expected;
}

int main(int argc, char** argv) {
   if (argc != 2) {
      printf("usage: %s <num>\n", argv[0]);
      printf("number of operations: <num> millions\n");
      exit(EXIT_FAILURE);
   }
   int n = atoi(argv[1]) * 1000000;
   if (n<=0)
       n=1000;

   double x = M_PI;
   double y = 1.0 + 1e-8;
   double t = stoptime();
   x = addmul(x, y, n);
   t = stoptime() - t;
   printf("addmul:\t %.3f s, %.3f Gflops, res=%f\n", t, (double)n/t/1e9, x);
   return EXIT_SUCCESS;
}

编译

g++ -O2 -march=native addmul.cpp ; ./a.out 1000

在英特尔酷睿i5-750,2.66 GHz上产生以下输出.

addmul:  0.270 s, 3.707 Gflops, res=1.326463

也就是说,每个周期只有大约1.4个触发器.用g++ -S -O2 -march=native -masm=intel addmul.cpp主循环查看汇编代码 对我来说似乎是最优的:

.L4:
inc    eax
mulsd    xmm8, xmm3
mulsd    xmm7, xmm3
mulsd    xmm6, xmm3
mulsd    xmm5, xmm3
mulsd    xmm1, xmm3
addsd    xmm13, xmm2
addsd    xmm12, xmm2
addsd    xmm11, xmm2
addsd    xmm10, xmm2
addsd    xmm9, xmm2
cmp    eax, ebx
jne    .L4

使用打包版本(addpd和mulpd)更改标量版本会使翻牌计数加倍,而不会改变执行时间,因此每个周期我只能获得2.8个翻牌.有一个简单的例子,每个周期实现四次触发吗？

Mysticial的精彩小程序; 这是我的结果(虽然运行了几秒钟):

• gcc -O2 -march=nocona:10.66 Gflops中的5.6 Gflops(2.1个翻牌/周期)
• cl /O2,openmp删除:10.1 Gflops 10.66 Gflops(3.8 flops/cycle)

这看起来有点复杂,但到目前为止我的结论:

• gcc -O2改变独立浮点运算的顺序,目的是交替 addpd,mulpd如果可能的话.同样适用于gcc-4.6.2 -O2 -march=core2.

• gcc -O2 -march=nocona 似乎保持了C++源代码中定义的浮点运算的顺序.

• cl /O2来自SDK for Windows 7的64位编译器会 自动进行循环展开,并且似乎尝试安排操作,以便三个组与三个组addpd交替mulpd(好吧,至少在我的系统和我的简单程序中) .

• My Core i5 750(Nahelem架构)不喜欢交替添加和mul,并且似乎无法并行运行这两个操作.但是,如果按3分组,它突然就像魔法一样.

• 其他架构(可能是Sandy Bridge和其他架构)似乎能够并行执行add/mul而不会出现问题,如果它们在汇编代码中交替出现.

• 虽然很难承认,但是我的系统在我的系统cl /O2的低级优化操作方面做得更好,并且为上面的小C++示例实现了接近峰值的性能.我在1.85-2.01翻牌/周期之间测量(在Windows中使用了clock()并不精确.我猜,需要使用更好的计时器 - 感谢Mackie Messer).

• 我管理的最好的gcc是手动循环展开并以三个为一组安排添加和乘法.随着 g++ -O2 -march=nocona addmul_unroll.cpp 我充其量0.207s, 4.825 Gflops只相当于1.8翻牌/周期,我现在很高兴.

在C++代码中,我用for循环替换了循环

   for (int i=0; i<loops/3; i++) {
       mul1*=mul; mul2*=mul; mul3*=mul;
       sum1+=add; sum2+=add; sum3+=add;
       mul4*=mul; mul5*=mul; mul1*=mul;
       sum4+=add; sum5+=add; sum1+=add;

       mul2*=mul; mul3*=mul; mul4*=mul;
       sum2+=add; sum3+=add; sum4+=add;
       mul5*=mul; mul1*=mul; mul2*=mul;
       sum5+=add; sum1+=add; sum2+=add;

       mul3*=mul; mul4*=mul; mul5*=mul;
       sum3+=add; sum4+=add; sum5+=add;
   }

现在装配看起来像

.L4:
mulsd    xmm8, xmm3
mulsd    xmm7, xmm3
mulsd    xmm6, xmm3
addsd    xmm13, xmm2
addsd    xmm12, xmm2
addsd    xmm11, xmm2
mulsd    xmm5, xmm3
mulsd    xmm1, xmm3
mulsd    xmm8, xmm3
addsd    xmm10, xmm2
addsd    xmm9, xmm2
addsd    xmm13, xmm2
...

Answer 1

我以前做过这个确切的任务.但主要是测量功耗和CPU温度.以下代码(相当长)在我的Core i7 2600K上实现了接近最佳状态.

这里需要注意的关键是大量的手动循环展开以及乘法和交错的交错......

完整的项目可以在我的GitHub上找到:https://github.com/Mysticial/Flops

警告:

如果您决定编译并运行它,请注意您的CPU温度!
确保不要让它过热.并确保CPU限制不会影响您的结果!

此外,对于运行此代码可能造成的任何损害,我不承担任何责任.

笔记:

• 此代码针对x64进行了优化.x86没有足够的寄存器来编译.
• 此代码已经过测试,可以在Visual Studio 2010/2012和GCC 4.6上正常运行.
  ICC 11(英特尔编译器11)令人惊讶地难以编译它.
• 这些适用于FMA之前的处理器.为了在Intel Haswell和AMD Bulldozer处理器(及更高版本)上实现峰值FLOPS,将需要FMA(融合乘法加法)指令.这些超出了本基准的范围.

#include <emmintrin.h>
#include <omp.h>
#include <iostream>
using namespace std;

typedef unsigned long long uint64;

double test_dp_mac_SSE(double x,double y,uint64 iterations){
    register __m128d r0,r1,r2,r3,r4,r5,r6,r7,r8,r9,rA,rB,rC,rD,rE,rF;

    //  Generate starting data.
    r0 = _mm_set1_pd(x);
    r1 = _mm_set1_pd(y);

    r8 = _mm_set1_pd(-0.0);

    r2 = _mm_xor_pd(r0,r8);
    r3 = _mm_or_pd(r0,r8);
    r4 = _mm_andnot_pd(r8,r0);
    r5 = _mm_mul_pd(r1,_mm_set1_pd(0.37796447300922722721));
    r6 = _mm_mul_pd(r1,_mm_set1_pd(0.24253562503633297352));
    r7 = _mm_mul_pd(r1,_mm_set1_pd(4.1231056256176605498));
    r8 = _mm_add_pd(r0,_mm_set1_pd(0.37796447300922722721));
    r9 = _mm_add_pd(r1,_mm_set1_pd(0.24253562503633297352));
    rA = _mm_sub_pd(r0,_mm_set1_pd(4.1231056256176605498));
    rB = _mm_sub_pd(r1,_mm_set1_pd(4.1231056256176605498));

    rC = _mm_set1_pd(1.4142135623730950488);
    rD = _mm_set1_pd(1.7320508075688772935);
    rE = _mm_set1_pd(0.57735026918962576451);
    rF = _mm_set1_pd(0.70710678118654752440);

    uint64 iMASK = 0x800fffffffffffffull;
    __m128d MASK = _mm_set1_pd(*(double*)&iMASK);
    __m128d vONE = _mm_set1_pd(1.0);

    uint64 c = 0;
    while (c < iterations){
        size_t i = 0;
        while (i < 1000){
            //  Here's the meat - the part that really matters.

            r0 = _mm_mul_pd(r0,rC);
            r1 = _mm_add_pd(r1,rD);
            r2 = _mm_mul_pd(r2,rE);
            r3 = _mm_sub_pd(r3,rF);
            r4 = _mm_mul_pd(r4,rC);
            r5 = _mm_add_pd(r5,rD);
            r6 = _mm_mul_pd(r6,rE);
            r7 = _mm_sub_pd(r7,rF);
            r8 = _mm_mul_pd(r8,rC);
            r9 = _mm_add_pd(r9,rD);
            rA = _mm_mul_pd(rA,rE);
            rB = _mm_sub_pd(rB,rF);

            r0 = _mm_add_pd(r0,rF);
            r1 = _mm_mul_pd(r1,rE);
            r2 = _mm_sub_pd(r2,rD);
            r3 = _mm_mul_pd(r3,rC);
            r4 = _mm_add_pd(r4,rF);
            r5 = _mm_mul_pd(r5,rE);
            r6 = _mm_sub_pd(r6,rD);
            r7 = _mm_mul_pd(r7,rC);
            r8 = _mm_add_pd(r8,rF);
            r9 = _mm_mul_pd(r9,rE);
            rA = _mm_sub_pd(rA,rD);
            rB = _mm_mul_pd(rB,rC);

            r0 = _mm_mul_pd(r0,rC);
            r1 = _mm_add_pd(r1,rD);
            r2 = _mm_mul_pd(r2,rE);
            r3 = _mm_sub_pd(r3,rF);
            r4 = _mm_mul_pd(r4,rC);
            r5 = _mm_add_pd(r5,rD);
            r6 = _mm_mul_pd(r6,rE);
            r7 = _mm_sub_pd(r7,rF);
            r8 = _mm_mul_pd(r8,rC);
            r9 = _mm_add_pd(r9,rD);
            rA = _mm_mul_pd(rA,rE);
            rB = _mm_sub_pd(rB,rF);

            r0 = _mm_add_pd(r0,rF);
            r1 = _mm_mul_pd(r1,rE);
            r2 = _mm_sub_pd(r2,rD);
            r3 = _mm_mul_pd(r3,rC);
            r4 = _mm_add_pd(r4,rF);
            r5 = _mm_mul_pd(r5,rE);
            r6 = _mm_sub_pd(r6,rD);
            r7 = _mm_mul_pd(r7,rC);
            r8 = _mm_add_pd(r8,rF);
            r9 = _mm_mul_pd(r9,rE);
            rA = _mm_sub_pd(rA,rD);
            rB = _mm_mul_pd(rB,rC);

            i++;
        }

        //  Need to renormalize to prevent denormal/overflow.
        r0 = _mm_and_pd(r0,MASK);
        r1 = _mm_and_pd(r1,MASK);
        r2 = _mm_and_pd(r2,MASK);
        r3 = _mm_and_pd(r3,MASK);
        r4 = _mm_and_pd(r4,MASK);
        r5 = _mm_and_pd(r5,MASK);
        r6 = _mm_and_pd(r6,MASK);
        r7 = _mm_and_pd(r7,MASK);
        r8 = _mm_and_pd(r8,MASK);
        r9 = _mm_and_pd(r9,MASK);
        rA = _mm_and_pd(rA,MASK);
        rB = _mm_and_pd(rB,MASK);
        r0 = _mm_or_pd(r0,vONE);
        r1 = _mm_or_pd(r1,vONE);
        r2 = _mm_or_pd(r2,vONE);
        r3 = _mm_or_pd(r3,vONE);
        r4 = _mm_or_pd(r4,vONE);
        r5 = _mm_or_pd(r5,vONE);
        r6 = _mm_or_pd(r6,vONE);
        r7 = _mm_or_pd(r7,vONE);
        r8 = _mm_or_pd(r8,vONE);
        r9 = _mm_or_pd(r9,vONE);
        rA = _mm_or_pd(rA,vONE);
        rB = _mm_or_pd(rB,vONE);

        c++;
    }

    r0 = _mm_add_pd(r0,r1);
    r2 = _mm_add_pd(r2,r3);
    r4 = _mm_add_pd(r4,r5);
    r6 = _mm_add_pd(r6,r7);
    r8 = _mm_add_pd(r8,r9);
    rA = _mm_add_pd(rA,rB);

    r0 = _mm_add_pd(r0,r2);
    r4 = _mm_add_pd(r4,r6);
    r8 = _mm_add_pd(r8,rA);

    r0 = _mm_add_pd(r0,r4);
    r0 = _mm_add_pd(r0,r8);


    //  Prevent Dead Code Elimination
    double out = 0;
    __m128d temp = r0;
    out += ((double*)&temp)[0];
    out += ((double*)&temp)[1];

    return out;
}

void test_dp_mac_SSE(int tds,uint64 iterations){

    double *sum = (double*)malloc(tds * sizeof(double));
    double start = omp_get_wtime();

#pragma omp parallel num_threads(tds)
    {
        double ret = test_dp_mac_SSE(1.1,2.1,iterations);
        sum[omp_get_thread_num()] = ret;
    }

    double secs = omp_get_wtime() - start;
    uint64 ops = 48 * 1000 * iterations * tds * 2;
    cout << "Seconds = " << secs << endl;
    cout << "FP Ops  = " << ops << endl;
    cout << "FLOPs   = " << ops / secs << endl;

    double out = 0;
    int c = 0;
    while (c < tds){
        out += sum[c++];
    }

    cout << "sum = " << out << endl;
    cout << endl;

    free(sum);
}

int main(){
    //  (threads, iterations)
    test_dp_mac_SSE(8,10000000);

    system("pause");
}

输出(1个线程,10000000次迭代) - 使用Visual Studio 2010 SP1编译 - x64版本:

Seconds = 55.5104
FP Ops  = 960000000000
FLOPs   = 1.7294e+010
sum = 2.22652

该机器是Core i7 2600K @ 4.4 GHz.理论SSE峰值为4个触发器*4.4 GHz = 17.6 GFlops.这段代码实现了17.3 GFlops - 不错.

输出(8个线程,10000000次迭代) - 使用Visual Studio 2010 SP1编译 - x64版本:

Seconds = 117.202
FP Ops  = 7680000000000
FLOPs   = 6.55279e+010
sum = 17.8122

理论SSE峰值为4个触发器*4个核心*4.4 GHz = 70.4个GFlops.实际是65.5 GFlops.

让我们更进一步.AVX ...

#include <immintrin.h>
#include <omp.h>
#include <iostream>
using namespace std;

typedef unsigned long long uint64;

double test_dp_mac_AVX(double x,double y,uint64 iterations){
    register __m256d r0,r1,r2,r3,r4,r5,r6,r7,r8,r9,rA,rB,rC,rD,rE,rF;

    //  Generate starting data.
    r0 = _mm256_set1_pd(x);
    r1 = _mm256_set1_pd(y);

    r8 = _mm256_set1_pd(-0.0);

    r2 = _mm256_xor_pd(r0,r8);
    r3 = _mm256_or_pd(r0,r8);
    r4 = _mm256_andnot_pd(r8,r0);
    r5 = _mm256_mul_pd(r1,_mm256_set1_pd(0.37796447300922722721));
    r6 = _mm256_mul_pd(r1,_mm256_set1_pd(0.24253562503633297352));
    r7 = _mm256_mul_pd(r1,_mm256_set1_pd(4.1231056256176605498));
    r8 = _mm256_add_pd(r0,_mm256_set1_pd(0.37796447300922722721));
    r9 = _mm256_add_pd(r1,_mm256_set1_pd(0.24253562503633297352));
    rA = _mm256_sub_pd(r0,_mm256_set1_pd(4.1231056256176605498));
    rB = _mm256_sub_pd(r1,_mm256_set1_pd(4.1231056256176605498));

    rC = _mm256_set1_pd(1.4142135623730950488);
    rD = _mm256_set1_pd(1.7320508075688772935);
    rE = _mm256_set1_pd(0.57735026918962576451);
    rF = _mm256_set1_pd(0.70710678118654752440);

    uint64 iMASK = 0x800fffffffffffffull;
    __m256d MASK = _mm256_set1_pd(*(double*)&iMASK);
    __m256d vONE = _mm256_set1_pd(1.0);

    uint64 c = 0;
    while (c < iterations){
        size_t i = 0;
        while (i < 1000){
            //  Here's the meat - the part that really matters.

            r0 = _mm256_mul_pd(r0,rC);
            r1 = _mm256_add_pd(r1,rD);
            r2 = _mm256_mul_pd(r2,rE);
            r3 = _mm256_sub_pd(r3,rF);
            r4 = _mm256_mul_pd(r4,rC);
            r5 = _mm256_add_pd(r5,rD);
            r6 = _mm256_mul_pd(r6,rE);
            r7 = _mm256_sub_pd(r7,rF);
            r8 = _mm256_mul_pd(r8,rC);
            r9 = _mm256_add_pd(r9,rD);
            rA = _mm256_mul_pd(rA,rE);
            rB = _mm256_sub_pd(rB,rF);

            r0 = _mm256_add_pd(r0,rF);
            r1 = _mm256_mul_pd(r1,rE);
            r2 = _mm256_sub_pd(r2,rD);
            r3 = _mm256_mul_pd(r3,rC);
            r4 = _mm256_add_pd(r4,rF);
            r5 = _mm256_mul_pd(r5,rE);
            r6 = _mm256_sub_pd(r6,rD);
            r7 = _mm256_mul_pd(r7,rC);
            r8 = _mm256_add_pd(r8,rF);
            r9 = _mm256_mul_pd(r9,rE);
            rA = _mm256_sub_pd(rA,rD);
            rB = _mm256_mul_pd(rB,rC);

            r0 = _mm256_mul_pd(r0,rC);
            r1 = _mm256_add_pd(r1,rD);
            r2 = _mm256_mul_pd(r2,rE);
            r3 = _mm256_sub_pd(r3,rF);
            r4 = _mm256_mul_pd(r4,rC);
            r5 = _mm256_add_pd(r5,rD);
            r6 = _mm256_mul_pd(r6,rE);
            r7 = _mm256_sub_pd(r7,rF);
            r8 = _mm256_mul_pd(r8,rC);
            r9 = _mm256_add_pd(r9,rD);
            rA = _mm256_mul_pd(rA,rE);
            rB = _mm256_sub_pd(rB,rF);

            r0 = _mm256_add_pd(r0,rF);
            r1 = _mm256_mul_pd(r1,rE);
            r2 = _mm256_sub_pd(r2,rD);
            r3 = _mm256_mul_pd(r3,rC);
            r4 = _mm256_add_pd(r4,rF);
            r5 = _mm256_mul_pd(r5,rE);
            r6 = _mm256_sub_pd(r6,rD);
            r7 = _mm256_mul_pd(r7,rC);
            r8 = _mm256_add_pd(r8,rF);
            r9 = _mm256_mul_pd(r9,rE);
            rA = _mm256_sub_pd(rA,rD);
            rB = _mm256_mul_pd(rB,rC);

            i++;
        }

        //  Need to renormalize to prevent denormal/overflow.
        r0 = _mm256_and_pd(r0,MASK);
        r1 = _mm256_and_pd(r1,MASK);
        r2 = _mm256_and_pd(r2,MASK);
        r3 = _mm256_and_pd(r3,MASK);
        r4 = _mm256_and_pd(r4,MASK);
        r5 = _mm256_and_pd(r5,MASK);
        r6 = _mm256_and_pd(r6,MASK);
        r7 = _mm256_and_pd(r7,MASK);
        r8 = _mm256_and_pd(r8,MASK);
        r9 = _mm256_and_pd(r9,MASK);
        rA = _mm256_and_pd(rA,MASK);
        rB = _mm256_and_pd(rB,MASK);
        r0 = _mm256_or_pd(r0,vONE);
        r1 = _mm256_or_pd(r1,vONE);
        r2 = _mm256_or_pd(r2,vONE);
        r3 = _mm256_or_pd(r3,vONE);
        r4 = _mm256_or_pd(r4,vONE);
        r5 = _mm256_or_pd(r5,vONE);
        r6 = _mm256_or_pd(r6,vONE);
        r7 = _mm256_or_pd(r7,vONE);
        r8 = _mm256_or_pd(r8,vONE);
        r9 = _mm256_or_pd(r9,vONE);
        rA = _mm256_or_pd(rA,vONE);
        rB = _mm256_or_pd(rB,vONE);

        c++;
    }

    r0 = _mm256_add_pd(r0,r1);
    r2 = _mm256_add_pd(r2,r3);
    r4 = _mm256_add_pd(r4,r5);
    r6 = _mm256_add_pd(r6,r7);
    r8 = _mm256_add_pd(r8,r9);
    rA = _mm256_add_pd(rA,rB);

    r0 = _mm256_add_pd(r0,r2);
    r4 = _mm256_add_pd(r4,r6);
    r8 = _mm256_add_pd(r8,rA);

    r0 = _mm256_add_pd(r0,r4);
    r0 = _mm256_add_pd(r0,r8);

    //  Prevent Dead Code Elimination
    double out = 0;
    __m256d temp = r0;
    out += ((double*)&temp)[0];
    out += ((double*)&temp)[1];
    out += ((double*)&temp)[2];
    out += ((double*)&temp)[3];

    return out;
}

void test_dp_mac_AVX(int tds,uint64 iterations){

    double *sum = (double*)malloc(tds * sizeof(double));
    double start = omp_get_wtime();

#pragma omp parallel num_threads(tds)
    {
        double ret = test_dp_mac_AVX(1.1,2.1,iterations);
        sum[omp_get_thread_num()] = ret;
    }

    double secs = omp_get_wtime() - start;
    uint64 ops = 48 * 1000 * iterations * tds * 4;
    cout << "Seconds = " << secs << endl;
    cout << "FP Ops  = " << ops << endl;
    cout << "FLOPs   = " << ops / secs << endl;

    double out = 0;
    int c = 0;
    while (c < tds){
        out += sum[c++];
    }

    cout << "sum = " << out << endl;
    cout << endl;

    free(sum);
}

int main(){
    //  (threads, iterations)
    test_dp_mac_AVX(8,10000000);

    system("pause");
}

输出(1个线程,10000000次迭代) - 使用Visual Studio 2010 SP1编译 - x64版本:

Seconds = 57.4679
FP Ops  = 1920000000000
FLOPs   = 3.34099e+010
sum = 4.45305

理论AVX峰值为8个触发器*4.4 GHz = 35.2个GFlops.实际是33.4 GFlops.

输出(8个线程,10000000次迭代) - 使用Visual Studio 2010 SP1编译 - x64版本:

Seconds = 111.119
FP Ops  = 15360000000000
FLOPs   = 1.3823e+011
sum = 35.6244

理论AVX峰值为8个触发器*4个核心*4.4 GHz = 140.8 GFlops.实际是138.2 GFlops.

现在进行一些解释:

性能关键部分显然是内循环内的48条指令.你会注意到它被分成4块,每块12条指令.这12个指令块中的每一个都完全相互独立 - 平均需要6个周期才能执行.

因此,在使用问题之间有12条指令和6个循环.乘法的延迟是5个周期,因此它足以避免延迟停顿.

需要规范化步骤以防止数据上溢/下溢.这是必需的,因为无操作代码将缓慢地增加/减少数据的大小.

因此,如果你只使用全零并且摆脱标准化步骤,那么它实际上可能做得比这更好.然而,由于我编写了测量功耗和温度的基准测试,我必须确保触发器是"真实"数据,而不是零 - 因为执行单元可能很好地为零使用更少功率的特殊情况处理并产生较少的热量.

更多结果:

• 英特尔酷睿i7 920 @ 3.5 GHz
• Windows 7旗舰版x64
• Visual Studio 2010 SP1 - x64发行版

主题:1

Seconds = 72.1116
FP Ops  = 960000000000
FLOPs   = 1.33127e+010
sum = 2.22652

理论SSE峰值:4个触发器*3.5 GHz = 14.0 GFlops.实际是13.3 GFlops.

主题:8

Seconds = 149.576
FP Ops  = 7680000000000
FLOPs   = 5.13452e+010
sum = 17.8122

理论SSE峰值:4个触发器*4个核心*3.5 GHz = 56.0 GFlops.实际是51.3 GFlops.

在多线程运行中,我的处理器温度达到了76C!如果运行这些,请确保结果不受CPU限制的影响.

• 2 x Intel Xeon X5482 Harpertown @ 3.2 GHz
• Ubuntu Linux 10 x64
• GCC 4.5.2 x64 - (-O2 -msse3 -fopenmp)

主题:1

Seconds = 78.3357
FP Ops  = 960000000000
FLOPs   = 1.22549e+10
sum = 2.22652

理论SSE峰值:4个触发器*3.2 GHz = 12.8个GFlops.实际是12.3 GFlops.

主题:8

Seconds = 78.4733
FP Ops  = 7680000000000
FLOPs   = 9.78676e+10
sum = 17.8122

理论SSE峰值:4个触发器*8个核心*3.2 GHz = 102.4 GFlops.实际值是97.9 GFlops.

Answer 2

人们经常忘记英特尔架构中的一点,调度端口在Int和FP/SIMD之间共享.这意味着在循环逻辑在浮点流中创建气泡之前,您将只获得一定数量的FP/SIMD突发.神秘主义者从他的代码中获得了更多的失败,因为他在展开的循环中使用了更长的步幅.

如果你看看Nehalem/Sandy Bridge架构 http://www.realworldtech.com/page.cfm?ArticleID=RWT091810191937&p=6, 那很清楚会发生什么.

相比之下,由于INT和FP/SIMD管道具有独立的发布端口和自己的调度程序,因此在AMD(Bulldozer)上应该更容易达到峰值性能.

这只是理论上的,因为我没有这些处理器来测试.

Answer 3

分支绝对可以阻止您维持峰值理论性能.如果您手动进行一些循环展开,您是否看到了差异？例如,如果每次循环迭代放置5到10倍的ops:

for(int i=0; i<loops/5; i++) {
      mul1*=mul; mul2*=mul; mul3*=mul; mul4*=mul; mul5*=mul;
      sum1+=add; sum2+=add; sum3+=add; sum4+=add; sum5+=add;
      mul1*=mul; mul2*=mul; mul3*=mul; mul4*=mul; mul5*=mul;
      sum1+=add; sum2+=add; sum3+=add; sum4+=add; sum5+=add;
      mul1*=mul; mul2*=mul; mul3*=mul; mul4*=mul; mul5*=mul;
      sum1+=add; sum2+=add; sum3+=add; sum4+=add; sum5+=add;
      mul1*=mul; mul2*=mul; mul3*=mul; mul4*=mul; mul5*=mul;
      sum1+=add; sum2+=add; sum3+=add; sum4+=add; sum5+=add;
      mul1*=mul; mul2*=mul; mul3*=mul; mul4*=mul; mul5*=mul;
      sum1+=add; sum2+=add; sum3+=add; sum4+=add; sum5+=add;
   }

Answer 4

在2.4GHz Intel Core 2 Duo上使用Intels icc Version 11.1

Macintosh:~ mackie$ icc -O3 -mssse3 -oaddmul addmul.cc && ./addmul 1000
addmul:  0.105 s, 9.525 Gflops, res=0.000000
Macintosh:~ mackie$ icc -v
Version 11.1 

这非常接近理想的9.6 Gflops.

编辑:

哎呀,看看汇编代码似乎icc不仅可以对乘法进行矢量化,还可以将循环中的附加物拉出来.强制更严格的fp语义,代码不再向量化:

Macintosh:~ mackie$ icc -O3 -mssse3 -oaddmul addmul.cc -fp-model precise && ./addmul 1000
addmul:  0.516 s, 1.938 Gflops, res=1.326463

EDIT2:

按照要求:

Macintosh:~ mackie$ clang -O3 -mssse3 -oaddmul addmul.cc && ./addmul 1000
addmul:  0.209 s, 4.786 Gflops, res=1.326463
Macintosh:~ mackie$ clang -v
Apple clang version 3.0 (tags/Apple/clang-211.10.1) (based on LLVM 3.0svn)
Target: x86_64-apple-darwin11.2.0
Thread model: posix

clang代码的内部循环如下所示:

        .align  4, 0x90
LBB2_4:                                 ## =>This Inner Loop Header: Depth=1
        addsd   %xmm2, %xmm3
        addsd   %xmm2, %xmm14
        addsd   %xmm2, %xmm5
        addsd   %xmm2, %xmm1
        addsd   %xmm2, %xmm4
        mulsd   %xmm2, %xmm0
        mulsd   %xmm2, %xmm6
        mulsd   %xmm2, %xmm7
        mulsd   %xmm2, %xmm11
        mulsd   %xmm2, %xmm13
        incl    %eax
        cmpl    %r14d, %eax
        jl      LBB2_4

EDIT3:

最后,有两点建议:首先,如果您喜欢这种类型的基准测试,请考虑使用该rdtsc指令gettimeofday(2).它更加准确并且可以提供循环时间,这通常是您感兴趣的.对于gcc和朋友,你可以这样定义:

#include <stdint.h>

static __inline__ uint64_t rdtsc(void)
{
        uint64_t rval;
        __asm__ volatile ("rdtsc" : "=A" (rval));
        return rval;
}

其次,您应该多次运行基准程序并仅使用最佳性能.在现代操作系统中,许多事情并行发生,cpu可能处于低频省电模式等.重复运行程序会给你一个更接近理想情况的结果.