c++ - 尝试将结构复制到设备内存时 CUDA 参数无效 (cudaMemcpy)

Question

我试图弄清楚我应该如何创建一个结构/类来发送到设备，但我一直收到这个“无效参数”CUDA 错误。我做了一个显示错误的小例子:

#include <iostream>
#include <cstdio>
using namespace std;

#define CUDA_WARN(XXX) \
    do { if (XXX != cudaSuccess) cerr << "CUDA Error: " << \
        cudaGetErrorString(XXX) << ", at line " << __LINE__ \
        << endl; cudaDeviceSynchronize(); } while (0)

struct P {
    double x,y;
    __host__ __device__ void init(const double &a, const double &b) {
        x = a; y = b; }
};

int main(int argc, char **argv)
{
    P hP, hQ, dP;
    cout << "Size of P: " << sizeof(P) << endl;
    CUDA_WARN(cudaMalloc((void**) &dP, sizeof(P)));
    printf("dP: %p\n", &dP); // print dP's address on the device
    hP.init(1.2,-2.1);
    hQ.init(0.,0.);
    CUDA_WARN(cudaMemcpy(&dP, &hP, sizeof(P), cudaMemcpyHostToDevice));
    CUDA_WARN(cudaMemcpy(&hQ, &dP, sizeof(P), cudaMemcpyDeviceToHost));
    cout << "Copy back: " << hQ.x << "\t" << hQ.y << endl;
    dP.init(3.,3.);
    CUDA_WARN(cudaMemcpy(&hP, &dP, sizeof(P), cudaMemcpyDeviceToHost));
    cout << "Copy new:  " << hP.x << "\t" << hP.y << endl;
    return 0;
}

我正在编译(我的卡是 Tesla C2050):

nvcc -arch sm_20 -o exec file.cu

我得到的结果是:

Size of P: 16
dP: 0x7fff82d4b7b0
CUDA Error: invalid argument, at line 24
CUDA Error: invalid argument, at line 25
Copy back: 0    0
CUDA Error: invalid argument, at line 28
Copy new:  1.2  -2.1


------------------
(program exited with code: 0)
Press return to continue

如果你能帮助我，谢谢你们!

====== 在@talonmies、@JackOLantern、@Robert Crovella 的评论之后 =======

谢谢大家!你真的帮了大忙!根据评论，我可以更正我的代码，现在它可以正常工作了。只是为了注册最终解决方案:

#include <iostream>
#include <cstdio>
using namespace std;

#define CUDA_WARN(XXX) \
    do { if (XXX != cudaSuccess) cerr << "CUDA Error: " << \
        cudaGetErrorString(XXX) << ", at line " << __LINE__ \
        << endl; cudaDeviceSynchronize(); } while (0)

struct P {
    double x,y;
    __host__ __device__ void init(const double &a, const double &b) {
        x = a; y = b; }
};

/* INCLUDED KERNEL FUNCTION */
__global__ void dev_P_init(P *p, double a, double b) {
    p->init(a,b);
}

int main(int argc, char **argv)
{
    P hP, hQ, *dP; //*changed*
    cout << "Size of P: " << sizeof(P) << endl;
    CUDA_WARN(cudaMalloc((void**) &dP, sizeof(P)));
    printf("dP: %p\n", &dP); // print dP's address on the device
    hP.init(1.2,-2.1);
    hQ.init(0.,0.);
    CUDA_WARN(cudaMemcpy(dP, &hP, sizeof(P), cudaMemcpyHostToDevice)); //*changed*
    CUDA_WARN(cudaMemcpy(&hQ, dP, sizeof(P), cudaMemcpyDeviceToHost)); //*changed*
    cout << "Copy back: " << hQ.x << "\t" << hQ.y << endl;
    dev_P_init<<< 1, 1 >>>(dP,3., 3.); //*call to kernel*
    CUDA_WARN(cudaMemcpy(&hP, dP, sizeof(P), cudaMemcpyDeviceToHost)); //*changed*
    cout << "Copy new:  " << hP.x << "\t" << hP.y << endl;
    return 0;
}

并更正后的输出:

Size of P: 16
dP: 0x7fff6fa2e498
Copy back: 1.2  -2.1
Copy new:  3    3


------------------
(program exited with code: 0)
Press return to continue

Answer 1

@talonmies 已经注意到，&dP 不是有效的设备指针。事实上，dP 是一个驻留在主机上的变量，因此它的地址指向主机内存空间。相反，当dP是一个指针时，cudaMalloc会接收它的值作为参数，它的值会指向一个设备内存空间。

这是您的代码的正确版本:

#include <iostream>
#include <cstdio>
using namespace std;

#define CUDA_WARN(XXX) \
    do { if (XXX != cudaSuccess) cerr << "CUDA Error: " << \
    cudaGetErrorString(XXX) << ", at line " << __LINE__ \
    << endl; cudaDeviceSynchronize(); } while (0)

struct P {
    double x,y;
    __host__ __device__ void init(const double &a, const double &b) {
    x = a; y = b; }
};

int main(int argc, char **argv)
{
    P *dP;
    P hP, hQ;
    CUDA_WARN(cudaMalloc((void**) &dP, sizeof(P)));
    CUDA_WARN(cudaMemcpy(dP, &hP, sizeof(P), cudaMemcpyHostToDevice));
    CUDA_WARN(cudaMemcpy(&hQ, dP, sizeof(P), cudaMemcpyDeviceToHost));
    CUDA_WARN(cudaMemcpy(&hP, dP, sizeof(P), cudaMemcpyDeviceToHost));

    return 0;
}