调用CUDA API使用GPU运算一个JuliaSet花费的时间比在CPU上多

Question

我用的是《GPU高性能编程CUDA实战》第四章最后的一个例子，在CPU上花费的时间700+ms，在GPU上花费的时间是800+ms

// CPU版本代码

void kernel(unsigned char * ptr) {
    for (size_t y = 0; y < DIM; y++) for (size_t x = 0; x < DIM; x++)
    {
        int offset = x + y*DIM;
        int juliaValue = julia(x, y);

        ptr[offset * 4 + 0] = 255 * juliaValue;
        ptr[offset * 4 + 1] = 0;
        ptr[offset * 4 + 2] = 0;
        ptr[offset * 4 + 3] = 255;
    }
}

int main(void) {

    clock_t start, finish;
    double totaltime;
    start = clock();

    CPUBitmap bitmap(DIM, DIM);
    unsigned char *ptr = bitmap.get_ptr();

    kernel(ptr);

    finish = clock();
    totaltime = (double)(finish - start) / CLOCKS_PER_SEC;
    printf("time : %fs\n", float(totaltime));

    bitmap.display_and_exit();

    return 0;
}

---

// GPU版本代码

#define DIM 1000

struct cuComplex
{
    float r;
    float i;
    __device__ cuComplex(float a, float b) : r(a), i(b) {}
    __device__ float magnitude2(void) { return r * r + i * i; }
    __device__ cuComplex operator*(const cuComplex& a) { return cuComplex(r*a.r - i*a.i, i*a.r + r*a.i); }
    __device__ cuComplex operator+(const cuComplex& a) { return cuComplex(r + a.r, i + a.i); }
};

__device__ int julia(int x, int y) {
    const float scale = 1.5;
    float jx = scale * (float)(DIM / 2 - x) / (DIM / 2);
    float jy = scale * (float)(DIM / 2 - y) / (DIM / 2);

    cuComplex c(-0.8, 0.156);
    cuComplex a(jx, jy);

    for (size_t i = 0; i < 200; i++)
    {
        a = a*a + c;
        if (a.magnitude2() > 1000) return 0;
    }

    return 1;
}

__global__ void kernel(unsigned char * ptr) {
    // 将threadIdx/BlockIdx映射到像素位置
    int x = blockIdx.x;
    int y = blockIdx.y;
    int offset = x + y*gridDim.x;

    // 计算对应位置上的值
    int juliaValue = julia(x, y);
    ptr[offset * 4 + 0] = 255 * juliaValue;
    ptr[offset * 4 + 1] = 0;
    ptr[offset * 4 + 2] = 0;
    ptr[offset * 4 + 3] = 255;
}

int main(void) {

    clock_t start, finish;
    double totaltime;
    start = clock();

    CPUBitmap bitmap(DIM, DIM);

    unsigned char *dev_bitmap;
    cudaMalloc((void **)&dev_bitmap, bitmap.image_size());

    dim3 grid(DIM, DIM);
    kernel<< <grid, 1 >> >(dev_bitmap);

    cudaMemcpy(bitmap.get_ptr(), dev_bitmap, bitmap.image_size(), cudaMemcpyDeviceToHost);

    cudaFree(dev_bitmap);

    finish = clock();
    totaltime = (double)(finish - start) / CLOCKS_PER_SEC;
    printf("time : %fs\n", float(totaltime));

    bitmap.display_and_exit();

    return 0;
}

运行结果是这张图片

熟悉CUDA编程的大神能否解释一下为什么使用GPU并行计算花费的时间反而跟多？

配置是i7-8700 + GTX1060 6G

Answer 1

julia是哪里的函数，对于gpu来说，调用函数开销很大，如果是调用主机上的函数，开销更大。