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嗨，我想实现一个推力非常大的循环，但我发现它比普通的 C++ 代码慢得多。你能告诉我我哪里出错了。fi 和 fj 是宿主向量

xsize 通常是一个 7-8 位数字

thrust::host_vector <double> df((2*floor(r)*(floor(r)+1)+1)*n*n); 
thrust::device_vector<double> gpu_df((2*floor(r)*(floor(r)+1)+1)*n*n); 
     for(i=0;i<xsize;i++)
     {
        gpu_df[i]=(fi[i]-fj[i]);

         if(gpu_df[i]<0)
            gpu_df[i]=0;
        else 
       gpu_df[i]=gpu_df[i]*(fi[i]-fj[i]);
        if(gpu_df[i]>255)
            gpu_df[i]=255;
        //      cout<<fi[i]<<"\n";
     }
df=gpu_df;

我觉得代码没有被并行化。你能帮帮我吗？

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1 回答 1

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要使用 Thrust 在 GPU 上运行程序，您需要根据 Thrust 算法（如reduce、transform、sort等）编写它们。在这种情况下，我们可以根据 来编写计算transform，因为循环只是计算一个函数F(fi[i], fj[i])并将结果存储在df[i]. 请注意，我们必须在调用之前先将输入数组移动到设备，transform因为 Thrust 要求输入和输出数组位于同一位置。

#include <thrust/host_vector.h>
#include <thrust/device_vector.h>
#include <thrust/functional.h>
#include <cstdio>

struct my_functor
  : public thrust::binary_function<float,float,float>
{
  __host__ __device__
  float operator()(float fi, float fj)
      {
    float d =  fi - fj;

    if (d < 0)
      d = 0;
    else
      d = d * d;

    if (d > 255)
      d = 255;

    return d;
  }
};

int main(void)
{
  size_t N = 5;

  // allocate storage on host
  thrust::host_vector<float>   cpu_fi(N);
  thrust::host_vector<float>   cpu_fj(N);
  thrust::host_vector<float>   cpu_df(N);

  // initialze fi and fj arrays
  cpu_fi[0] = 2.0;  cpu_fj[0] =  0.0;
  cpu_fi[1] = 0.0;  cpu_fj[1] =  2.0;
  cpu_fi[2] = 3.0;  cpu_fj[2] =  1.0;
  cpu_fi[3] = 4.0;  cpu_fj[3] =  5.0;
  cpu_fi[4] = 8.0;  cpu_fj[4] = -8.0;

  // copy fi and fj to device
  thrust::device_vector<float> gpu_fi = cpu_fi;
  thrust::device_vector<float> gpu_fj = cpu_fj;

  // allocate storage for df
  thrust::device_vector<float> gpu_df(N);

  // perform transformation
  thrust::transform(gpu_fi.begin(), gpu_fi.end(),  // first input range
                    gpu_fj.begin(),                // second input range
                    gpu_df.begin(),                // output range
                    my_functor());                 // functor to apply

  // copy results back to host
  thrust::copy(gpu_df.begin(), gpu_df.end(), cpu_df.begin());

  // print results on host
  for (size_t i = 0; i < N; i++)
    printf("f(%2.0lf,%2.0lf) = %3.0lf\n", cpu_fi[i], cpu_fj[i], cpu_df[i]);

  return 0;
}

For reference, here's the output of the program:

f( 2, 0) =   4
f( 0, 2) =   0
f( 3, 1) =   4
f( 4, 5) =   0
f( 8,-8) = 255
于 2011-06-15T03:11:15.867 回答