Reputation: 25
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.
Upvotes: 2
Views: 2923
Reputation: 22271
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.
Upvotes: 0
Reputation: 1176
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.
Upvotes: 1