cudaMemcpy2D: What's wrong in this code?

Accelerated Computing CUDA CUDA Programming and Performance
1

Hi everybody

I’m trying to pass a 2D matrix of 0’s to device to calculate in the gpu a simple operation:

for example if I have a matrix[height][width] where heigh = 2 an width = 2

matrix[0][0] = 0

matrix[0][1] = 1

matrix[1][0] = 2

matrix[1][1] = 3

Finally I want copy the results from device to host and print results. The code works fine if height = widht, but it doesn’t work if height != width

Any Ideas?

Thank you. I’m desperate :(

#include <stdio.h>

#include <stdlib.h>

#include <cuda.h>

#include <curand_kernel.h>

__global__ void MyKernel(float** dev_matrix, size_t pitch, int width, int height)

{   

	int number = 0;

	for (int i = 0; i < height; ++i) 

	{

		float* row = (float*)((char*)dev_matrix + i*pitch);

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

		{	

			row[j] = number; 

			number++;

		}

	}

}

int main (int argc , char * argv [])

{

	int width = 4, height = 2, i, j;

	float matrix[width][height];

	float **dev_matrix;

	size_t pitch;

printf("\nMATRIX MANIPULATION\n");

	for (i = 0; i < height; i++) 

		for (j = 0; j < width; j++)

			matrix[i][j] = 0.0;

	printf("Matrix in host memory\n");

	for (i = 0; i < height; i++) 

	{	

		for (j = 0; j < width; j++)

			printf("%f   ", matrix[i][j]);

		printf("\n");

	}

	cudaMallocPitch(&dev_matrix, &pitch, width * sizeof(float), height);

	cudaMemcpy2D(dev_matrix, pitch, matrix, width * sizeof(float), width * sizeof(float), height, cudaMemcpyHostToDevice);  

	MyKernel<<<1, 1>>>(dev_matrix, pitch, width, height);

	cudaMemcpy2D(matrix, width * sizeof(float), dev_matrix, pitch, width * sizeof(float), height, cudaMemcpyDeviceToHost);

	printf("Matrix after calculate elements in the gpu\n");

	for (i = 0; i < height; i++) 

	{	

		for (j = 0; j < width; j++)

			printf("%f   ", matrix[i][j]);

		printf("\n");

	}

cudaFree(dev_matrix);

	return 0;

}
2

any ideas?

3

If you want to move only 0’s to the device you do not need a memory copy operation, you need a memory set operation, which is much faster.

If you are 100% sure each element is processed you do not even need a memory set operation, the allocation is just enough, since you write the output of every element.

About memCpy2D i cannot help because I never used 2D and 3D before since the same can be realized with simple memCpy and an array of pointer pointing to the individual rows

4

Thank you for the reply

This is an example. In the real code I move random numbers from the host to device.

I think the problem is in the CudaMemcpy2D. More ideas, please?

5 
float matrix[width][height];

but

for (i = 0; i < height; i++) 

                for (j = 0; j < width; j++)

                        matrix[i][j] = 0.0;

printf("Matrix in host memory\n");

        for (i = 0; i < height; i++) 

        {       

                for (j = 0; j < width; j++)

                        printf("%f   ", matrix[i][j]);

                printf("\n");

        }

I think you need to reverse your width and height in the loop or in the declaration

6

Oh my god…thank you so much.
English is not my languaje, and I’ve confused wiht width and height.
Thanks again!