sending struct array to cuda kernel

Question

I'm working on a project and I have to sent a struct array to cuda kernel. The struct also contains an array. To test it I have written a simple program.

struct Point {
    short     x;
    short     *y;
};

my kernel code:

__global__ void addKernel(Point *a, Point *b, Point *c)
{
    int i = threadIdx.x;

    c[i].x = a[i].x + b[i].x;
    for (int j = 0; j<4; j++){
        c[i].y[j] = a[i].y[j] + a[i].y[j];
    }
}

my main code:

int main()
{
    const int arraySize = 4;
    const int arraySize2 = 4;

    short *ya, *yb, *yc;
    short *dev_ya, *dev_yb, *dev_yc;

    Point *a;
    Point *b;
    Point *c;
    Point *dev_a;
    Point *dev_b;
    Point *dev_c;

    size_t sizeInside = sizeof(short) * arraySize2;

    ya = (short *)malloc(sizeof(short) * arraySize2);
    yb = (short *)malloc(sizeof(short) * arraySize2);
    yc = (short *)malloc(sizeof(short) * arraySize2);

    ya[0] = 1; ya[1] =2; ya[2]=3; ya[3]=4;
    yb[0] = 2; yb[1] =3; yb[2]=4; yb[3]=5;

    size_t sizeGeneral = (sizeInside+sizeof(short)) * arraySize;

    a = (Point *)malloc( sizeGeneral );  
    b = (Point *)malloc( sizeGeneral );
    c = (Point *)malloc( sizeGeneral );


    a[0].x = 2;  a[0].y = ya;
    a[1].x = 2;  a[1].y = ya;
    a[2].x = 2;  a[2].y = ya;
    a[3].x = 2;  a[3].y = ya;

    b[0].x = 4;  b[0].y = yb;
    b[1].x = 4;  b[1].y = yb;
    b[2].x = 4;  b[2].y = yb;
    b[3].x = 4;  b[3].y = yb;

    cudaMalloc((void**)&dev_a, sizeGeneral);
    cudaMalloc((void**)&dev_b, sizeGeneral);
    cudaMalloc((void**)&dev_c, sizeGeneral);

    cudaMemcpy(dev_a, a, sizeGeneral, cudaMemcpyHostToDevice);
    cudaMemcpy(dev_b, b, sizeGeneral, cudaMemcpyHostToDevice);

    addKernel<<<1, 4>>>(dev_a, dev_b, dev_c);

    cudaError_t err = cudaMemcpy(c, dev_c, sizeGeneral, cudaMemcpyDeviceToHost);   

    printf("{%d-->%d,%d,%d,%d} \n err= %d",c[0].x,c[0].y[0],c[1].y[1],c[1].y[2],c[2].y[3], err);        

    cudaFree(dev_a);
    cudaFree(dev_b);
    cudaFree(dev_c);

    return 0;
}

It seems cuda kernel is not working. Actually I can access structs 'x' variable but I cannot access 'y' array. What can I do to access the 'y' array? Thanks in advance.

What kind of error occurs? What does your application output? — Dariusz
– Dariusz, Commented Apr 12, 2013 at 7:49
it has not a compile error. In kernel cannot access 'y' so, when i want to access 'c[i].y[j]', it crash. — Soner İlica
– Soner İlica, Commented Apr 12, 2013 at 7:54

Adrian · Accepted Answer · 2013-04-12 07:41:22Z

1

When you are sending this struct to kernel you send short and pointer to short in host memory not device. This is crucial. For simple type - as short this works, because kernel has its local copy in memory designated to accept parameters. So when you call this kernel you have moved x and y to device, but not the area pointed by y. This you have to do manually by allocating space for it and updating pointer y to point to device memory.

answered Apr 12, 2013 at 7:41

Adrian

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Comments

Dariusz · Accepted Answer · 2013-04-12 07:51:32Z

0

You are not passin the array to the device. You can either make the array a part of the struct, by defining it like this:

struct {
  short normalVal;
  short inStructArr[4];
}

Or pass the array into the device memory and update the pointer in the struct.

answered Apr 12, 2013 at 7:51

Dariusz

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1 Comment

Robert Crovella Over a year ago

Handling pointers to device memory inside of data structures that reside in device memory is somewhat complicated. The modified struct method proposed here is much simpler. If you insist on passing the pointer, as Adrian indicates in his answer, then you need to use a method similar to what I described here. Nevertheless I don't recommend it due to it's complexity, if you can avoid it.

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