2D array of int copy from host to device and vice versa How to do it?

Accelerated Computing CUDA CUDA Programming and Performance
1

Hi, I’ve to copy an array of int from the host to device and tried to use the functions cudaMallocPitch and cudaMemcpy2D but all I obtained is a segmentation fault when I try to modify these data on the device…

This is the code that I wrote:

on the host:

int** vect=new int*[20];

	for (int i=0; i<20; ++i) {

		vect[i]=new int[30];

		for (int j=0; j<30; ++j)

			vect[i][j]=i*j;

	}

		

	int** d_vect;

	size_t pitch=sizeof(int);

	cudaMallocPitch((void**) &d_vect, &pitch, 30*sizeof(int), 20);

	cudaMemcpy2D(d_vect, pitch, vect, pitch, 30*sizeof(int), 20, cudaMemcpyHostToDevice);

	dim3 grid(30,20);

	nothing<<< grid, 1 >>>(d_vect);

	cudaMemcpy2D(vect, pitch, d_vect, pitch, 30*sizeof(int), 20, cudaMemcpyDeviceToHost);

__global__ void nothing(int** vect) {

	vect[blockIdx.x][blockIdx.y]=-vect[blockIdx.x][blockIdx.y];

}

I’m sure there’re some error in this code, can you explain me where these are?

I’m not sure I understood well how to do… Many thanks and excuse me for the disturb but it’s my first code with cuda…

2

The error is in the level of indirection:

cudaMallocPitch() does allocate a “flat” array of pitchheigth bytes (use int instead of int**).