How to copy variables from a custom class array on host into a float array on device in CUDA

Question

I am using CUDA. I have the following class on host:

class Particle{
     public:
     float x;
     float v;
     // several other variables
}

Then I have a vector of particles

vector <Particle> p_all(512);

On the GPU, I want to operate on an array of all x's (taken from all the Particles), and want to copy the data from the Particles array into a float array on device. I have a hunch that cudaMemcpy can be used, and I tried the following code, but it gives invalid pitch error.

cudaMalloc( (void**) &pos_dev, sizeof(float)*512);
cudaMemcpy2D( (void*) &pos_dev, sizeof(float), (void*)&p_all[0].x, sizeof(Particle), sizeof(Particle), 512*sizeof(float), cudaMemcpyHostToDevice);

Is it at all possible to do so? Of course, the backup solution is to create an array of x's using a for loop and then copy it to the device. But I am looking for a more efficient solution.

Thanks.

FULL CODE BELOW.

#include <cuda_runtime.h>
#include <iostream>
#include <vector>
using namespace std;

// This will output the proper error string when calling cudaGetLastError
void getLastCudaError(string s=""){
    string errMessage = s;
    cudaError_t err = cudaGetLastError();
    if( err != cudaSuccess){
        cerr << __FILE__ << "(" << __LINE__ << ") : Last Cuda Error - " << errMessage 
             << " (" << int(err) << "): " << cudaGetErrorString(err) << ".\n";
        exit(-1);
    }
}

class Particle{
    public:
    float x;
    float v;
    int a;
    char c;
    short b;

    Particle(){
        a=1988; c='a'; v=5.56; x=1810; b=1.66;
    }
};

template <class T>
void printVec(vector <T> &v, string name = "v"){
    cout << name << " = ";
    for (int i=0; i<v.size(); ++i) cout << v[i] << " " ;
    cout << '\n';
}

int main(){

    const int N = 512;
    vector <float> pos(N,5);

    vector <Particle> p_all(N);

    float * pos_dev;
    float * vel_dev;

    cudaMalloc( (void**) &pos_dev, sizeof(float)*N);

    printVec(pos, "pos");

    cudaMemcpy2D( (void*) &pos_dev, sizeof(float), (void*)&(p_all[0].x), sizeof(Particle), sizeof(float), N, cudaMemcpyHostToDevice);
    getLastCudaError("HtoD");

    cudaMemcpy( (void*) &pos[0], (void*)&pos_dev, N*sizeof(float), cudaMemcpyDeviceToHost);
    getLastCudaError("DtoH");

    printVec(pos, "pos_new");

    return 0;

}

Answer 1

You are allocating your data as "array of structures", like

class Particle{
    public:
        float x;
        float v;
}

Particle foo[N];

which will lead to coalescing issues due to the data interleaving and for this reason you are trying to use cudaMemcpy2D . A more convenient solution in terms of bandwidth exploitation is allocating the data as "structures of arrays" as

class Particle{
    public:
        float x[N];
        float v[N];
}

Particle foo;

In this way, you will be able to avoid the use of cudaMemcpy2D and copy the data from host to device by a simple cudaMemcpy .

Answer 2

Your cudaMemcpy2D call is set up incorrectly. Check the documentation .

try this instead:

cudaMemcpy2D( (void*) pos_dev, sizeof(float), (void*)&(p_all[0].x), sizeof(Particle), sizeof(float), 512, cudaMemcpyHostToDevice);

There were multiple parameters that needed to be modified, but the invalid pitch error came about because the requested width of transfer in bytes (you had sizeof(Particle) ) was wider than the destination pitch ( sizeof(float) , which is correct)

EDIT: in addition, although you didn't ask about it, the final cudaMemcpy operation in the code you have now posted is also incorrect. The following changes should help:

cudaMemcpy( (void*) &(pos[0]), (void*)pos_dev, N*sizeof(float), cudaMemcpyDeviceToHost);