c - 如何在 C 语言中改进/加速此频率函数？

Question

我怎样才能改进/加快这个频繁使用的功能？

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

#define M 10 // This is fixed
#define N 8  // This is NOT fixed

// Assumptions: 1. x, a, b and c are all arrays of 10 (M).
//              2. y and z are all matrices of 8 x 10 (N x M).
// Requirement: 1. return the value of ret;
//              2. get all elements of array c
float fnFrequentFunction(const float* x, const float* const* y, const float* const* z,
                         const float* a, const float* b, float *c, int n)
{
    register float tmp;
    register float sum;
    register float ret = 0;
    register const float* yy;
    register const float* zz;
    int i;

    for (i = 0; i < n; i++)  // M == 1, 2, 4, or 8
    {
        sum = 0;
        yy = y[i];
        zz = z[i];

        tmp = x[0] - yy[0]; sum += tmp * tmp * zz[0];
        tmp = x[1] - yy[1]; sum += tmp * tmp * zz[1];
        tmp = x[2] - yy[2]; sum += tmp * tmp * zz[2];
        tmp = x[3] - yy[3]; sum += tmp * tmp * zz[3];
        tmp = x[4] - yy[4]; sum += tmp * tmp * zz[4];
        tmp = x[5] - yy[5]; sum += tmp * tmp * zz[5];
        tmp = x[6] - yy[6]; sum += tmp * tmp * zz[6];
        tmp = x[7] - yy[7]; sum += tmp * tmp * zz[7];
        tmp = x[8] - yy[8]; sum += tmp * tmp * zz[8];
        tmp = x[9] - yy[9]; sum += tmp * tmp * zz[9];

        ret += (c[i] = log(a[i] * b[i]) + sum);
    }

    return ret;
}

// In the main function, all values are just example data.
int main()
{
    float x[M] = {0.001251f, 0.563585f, 0.193304f, 0.808741f, 0.585009f, 0.479873f, 0.350291f, 0.895962f, 0.622840f, 0.746605f};
    float* y[N];
    float* z[N];
    float a[M] = {0.870205f, 0.733879f, 0.711386f, 0.588244f, 0.484176f, 0.852962f, 0.168126f, 0.684286f, 0.072573f, 0.632160f};
    float b[M] = {0.871487f, 0.998108f, 0.798608f, 0.134831f, 0.576281f, 0.410779f, 0.402936f, 0.522935f, 0.623218f, 0.193030f};
    float c[N];

    float t1[M] = {0.864406f, 0.709006f, 0.091433f, 0.995727f, 0.227180f, 0.902585f, 0.659047f, 0.865627f, 0.846767f, 0.514359f};
    float t2[M] = {0.866817f, 0.581347f, 0.175542f, 0.620197f, 0.781823f, 0.778588f, 0.938688f, 0.721610f, 0.940214f, 0.811353f};
    int i, j;

    int n = 10000000;
    long start;

    // Initialize y, z for test example:
    for(i = 0; i < N; ++i)
    {
        y[i] = (float*)malloc(sizeof(float) * M);
        z[i] = (float*)malloc(sizeof(float) * M);

        for(j = 0; j < M; ++j)
        {
            y[i][j] = t1[j] * j;
            z[i][j] = t2[j] * j;
        }
    }


    // Speed test here:
    start = clock();
    while(--n)
        fnFrequentFunction(x, y, z, a, b, c, 8);
    printf("Time used: %ld\n", clock() - start);


    // Output the result here:
    printf("fnFrequentFunction == %f\n", fnFrequentFunction(x, y, z, a, b, c, 8));
    for(j = 0; j < N; ++j)
        printf("  c[%d] == %f\n", j, c[j]);
    printf("\n");


    // Free memory
    for(j = 0; j < N; ++j)
    {
        free(y[j]);
        free(z[j]);
    }

    return 0;
}

欢迎提出任何建议:-)

我感到很糟糕，因为我在我的函数中犯了一个大错误。上面的代码是新的。我现在正在重新检查以确保这是我需要的。

Answer 1

把它放在循环之外

sum = 0;

tmp = x[0] - y[0]; sum += tmp * tmp * z[0];
tmp = x[1] - y[1]; sum += tmp * tmp * z[1];
tmp = x[2] - y[2]; sum += tmp * tmp * z[2];
tmp = x[3] - y[3]; sum += tmp * tmp * z[3];
tmp = x[4] - y[4]; sum += tmp * tmp * z[4];
tmp = x[5] - y[5]; sum += tmp * tmp * z[5];
tmp = x[6] - y[6]; sum += tmp * tmp * z[6];
tmp = x[7] - y[7]; sum += tmp * tmp * z[7];
tmp = x[8] - y[8]; sum += tmp * tmp * z[8];
tmp = x[9] - y[9]; sum += tmp * tmp * z[9];