Question

我有以下（可编译和可执行的）代码，使用CUDA Thrust来执行float2数组的减少。它工作正常

using namespace std;

// includes, system 
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <conio.h>

#include <typeinfo>  
#include <iostream>

// includes CUDA
#include <cuda.h>
#include <cuda_runtime.h>

// includes Thrust
#include <thrust/host_vector.h>
#include <thrust/device_vector.h>
#include <thrust/reduce.h>

// float2 + struct
struct add_float2 {
    __device__ float2 operator()(const float2& a, const float2& b) const {
        float2 r;
        r.x = a.x + b.x;
        r.y = a.y + b.y;
        return r;
    }
 };

// double2 + struct
struct add_double2 {
    __device__ double2 operator()(const double2& a, const double2& b) const {
        double2 r;
        r.x = a.x + b.x;
        r.y = a.y + b.y;
        return r;
    }
 };

void main( int argc, char** argv) 
{
    int N = 20;

    // --- Host
    float2* ha; ha = (float2*) malloc(N*sizeof(float2));
    for (unsigned i=0; i<N; ++i) {
        ha[i].x = 1;
        ha[i].y = 2;
    }

    // --- Device
    float2* da; cudaMalloc((void**)&da,N*sizeof(float2));
    cudaMemcpy(da,ha,N*sizeof(float2),cudaMemcpyHostToDevice);

    thrust::device_ptr<float2> dev_ptr_1(da);
    thrust::device_ptr<float2> dev_ptr_2(da+N);

    float2 init; init.x = init.y = 0.0f;

    float2 sum = thrust::reduce(dev_ptr_1,dev_ptr_2,init,add_float2());

    cout << " Real part = " << sum.x << "; Imaginary part = " << sum.y << endl;

    getch();

 }

但是，当我在float2计划中将double2更改为main时，即

void main( int argc, char** argv) 
{
    int N = 20;

    // --- Host
    double2* ha; ha = (double2*) malloc(N*sizeof(double2));
    for (unsigned i=0; i<N; ++i) {
        ha[i].x = 1;
        ha[i].y = 2;
    }

    // --- Device
    double2* da; cudaMalloc((void**)&da,N*sizeof(double2));
    cudaMemcpy(da,ha,N*sizeof(double2),cudaMemcpyHostToDevice);

    thrust::device_ptr<double2> dev_ptr_1(da);
    thrust::device_ptr<double2> dev_ptr_2(da+N);

    double2 init; init.x = init.y = 0.0;

    double2 sum = thrust::reduce(dev_ptr_1,dev_ptr_2,init,add_double2());

    cout << " Real part = " << sum.x << "; Imaginary part = " << sum.y << endl;

    getch();

}

我在exception行收到reduce。如何在double2阵列中使用CUDA推力减少？我做错了吗？提前谢谢。

TALONMIES答案后的工作解决方案

使用namespace std;

// includes, system
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <conio.h>

#include <typeinfo>
#include <iostream>

// includes CUDA
#include <cuda.h>
#include <cuda_runtime.h>

// includes Thrust
#include <thrust/host_vector.h>
#include <thrust/device_vector.h>
#include <thrust/reduce.h>

struct my_double2 {
    double x, y;
};

// double2 + struct
struct add_my_double2 {
    __device__ my_double2 operator()(const my_double2& a, const my_double2& b) const {
        my_double2 r;
        r.x = a.x + b.x;
        r.y = a.y + b.y;
        return r;
    }
};

void main( int argc, char** argv) 
{
    int N = 20;

    // --- Host
    my_double2* ha; ha = (my_double2*) malloc(N*sizeof(my_double2));
    for (unsigned i=0; i<N; ++i) {
        ha[i].x = 1;
        ha[i].y = 2;
    }

    // --- Device
    my_double2* da; cudaMalloc((void**)&da,N*sizeof(my_double2));
    cudaMemcpy(da,ha,N*sizeof(my_double2),cudaMemcpyHostToDevice);

    thrust::device_ptr<my_double2> dev_ptr_1(da);
    thrust::device_ptr<my_double2> dev_ptr_2(da+N);

    my_double2 init; init.x = init.y = 0.0;

    cout << "here3\n";
    my_double2 sum = thrust::reduce(dev_ptr_1,dev_ptr_2,init,add_my_double2());

    cout << " Real part = " << sum.x << "; Imaginary part = " << sum.y << endl;

    getch();

}

Answer 1

这是与MSVC和nvcc的已知不兼容性。例如，请参阅here。解决方案是定义您自己的double2版本并使用它。

仅供参考，我可以使用CUDA 5.5在Linux 64位盒上正确编译和运行代码。

CUDA使用double2阵列减少推力

1 个答案: