使用CUDA的Thrust库获取较大的值
|
嗨，我想实现一个推力极大的循环，但是我发现它比普通的C ++代码慢得多。你能告诉我我要去哪里错了。
fi和fj是宿主载体
xsize通常是大约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;
没有找到相关结果
已邀请:
1 个回复
谷起
，，等。在这种情况下，我们可以用编写计算，因为循环只是计算函数和将结果存储在“ 6”中。请注意，在调用之前，我们必须先将输入数组移至设备，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; }