在__host____device__functor中创建Thrust::device_vectors

我目前正在尝试并行化当前在主函数中顺序运行的thrust-cuda代码（因此不利用GPU的功能）。我基本上已经获取了函数代码，并将其放入一个函数子中，该函数子可以使用cuda流调用thrust:：for_each。然而，如果我使用定义函子

__host__ __device__ 

VS2013抛出各种警告，说我正试图从设备启动主机功能。这些错误发生在我使用定义矢量的地方

thrust::device_vector vect (size_vector); 

以及一些推力：：转换函数。它特别引用了thrust:：device_malloc_allocater的问题。如果我将函子严格定义为主机函子，这些错误都会消失，但当我使用探查器时，很明显，只有0.01%的设备被使用，这让我相信for_each实际上并没有在函子中启动推力代码。

编辑下面是一些编译并显示此错误的代码

#include <iostream>
#include <thrust/device_vector.h>
#include <thrust/host_vector.h>
#include <thrust/sort.h>
#include <thrust/execution_policy.h>
#include <thrust/for_each.h>
#include <thrust/sequence.h>
#include <cstdlib>
#include <ctime>
#include <vector>
#include <algorithm>
#include <memory.h>
#include <cstdio>
#include <thread>
#include <thrust/copy.h>
#include <thrust/iterator/zip_iterator.h>
#include <thrust/reduce.h>

using namespace std;
const int num_segs = 1;  // number of segments to sort
const int num_vals = 5;  // number of values in each segment

template <typename T> 
struct sort_vector
{
    T *Ddata;
    T *vect3;
    T *answer;
    sort_vector(T *_Ddata, T *_vect3, float *a) : Ddata(_Ddata), vect3(_vect3), answer(a) {};

    __host__ __device__ void operator()(int idx)
    {
        thrust::sort(thrust::seq, Ddata + idx*num_vals, Ddata + ((idx + 1)*num_vals));
        thrust::device_ptr<float> vect3_ptr = thrust::device_pointer_cast(vect3);
        thrust::device_vector<float> vect(10, 1);
        thrust::device_vector<float> vect2(10, 3);
        thrust::transform(thrust::device, vect.begin(), vect.end(), vect2.begin(), vect3_ptr, thrust::minus<float>());
        *answer = thrust::reduce(thrust::device, Ddata + idx*num_vals, Ddata + ((idx + 1)*num_vals));
    }
};
int main() {
    thrust::device_vector<float> d_Ddata(num_segs*num_vals);
    d_Ddata[0] = 50;
    d_Ddata[1] = 9.5;
    d_Ddata[2] = 30;
    d_Ddata[3] = 8.1;
    d_Ddata[4] = 1;
    thrust::device_vector<float> d_Ddata2(num_segs*num_vals);
    d_Ddata2[0] = 50;
    d_Ddata2[1] = 20.5;
    d_Ddata2[2] = 70;
    d_Ddata2[3] = 8.1;
    d_Ddata2[4] = 1;
    thrust::device_vector<float> vect3(10, 0);
    thrust::device_vector<float> vect4(10, 0);
    cout << "original dut" << endl;
    int g = 0;
        while (g < num_segs*num_vals){
            cout << d_Ddata[g] << endl;
            g++;
        }
        thrust::device_vector<int> d_idxs(num_segs);
        thrust::sequence(d_idxs.begin(), d_idxs.end());
        thrust::device_vector<float> dv_answer(1);
        thrust::device_vector<float> dv_answer2(1);
        cudaStream_t s1, s2;
        cudaStreamCreate(&s1);
        cudaStreamCreate(&s2);
        clock_t start;
        double duration;
        start = clock();
        thrust::for_each(thrust::cuda::par.on(s1),
            d_idxs.begin(),
            d_idxs.end(), sort_vector<float>(thrust::raw_pointer_cast(d_Ddata.data()), thrust::raw_pointer_cast(vect3.data()), thrust::raw_pointer_cast(dv_answer.data())));
        thrust::for_each(thrust::cuda::par.on(s2),
            d_idxs.begin(),
            d_idxs.end(), sort_vector<float>(thrust::raw_pointer_cast(d_Ddata2.data()), thrust::raw_pointer_cast(vect4.data()), thrust::raw_pointer_cast(dv_answer2.data())));
        cudaStreamSynchronize(s1);
        cudaStreamSynchronize(s2);
        cout << "sorted dut" << endl;
        int n = 0;
        while (n < num_segs*num_vals){
            cout << d_Ddata[n] << endl;
            n++;
        } 
        cout << "sum" << endl;
        cout << dv_answer[0] << endl;
        cout << dv_answer2[0] << endl;
        cout << "vector subtraction" << endl;
        int e = 0;
        while (e < 10){
            cout << vect3[e] << endl;
            e++;
        }
        cudaStreamDestroy(s1);
        cudaStreamDestroy(s2);
        duration = (clock() - start) / (double)CLOCKS_PER_SEC;
        cout << "time " << duration << endl;
        cin.get();
        return 0;
    }

thrust:：for_each是否可能不能调用__host__函子？

一些推力呼叫是否与幕后主持人有着天生的联系？

我能看到的唯一可能的解决方法是创建一个__host__ __device__函数库，该函数库中有单独的主机和设备定义代码。也有可能是我在研究这个主题时遗漏了一些东西。如有任何建议，我们将不胜感激。