float2CUDA Thrust を使用して配列のリダクションを実行する次の (コンパイル可能で実行可能な) コードがあります。正しく動作します
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();
}
ただし、プログラムでに変更すると、float2つまりdouble2main
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();
}
exceptionラインで受け取りますreduce。double2配列で CUDA Thrust リダクションを使用するにはどうすればよいですか? 私は何か間違ったことをしていますか？前もって感謝します。
TALONMIES の回答に続く実用的なソリューション
名前空間 std を使用します。
// 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();
}