Allocate 2D Array on Device Memory in CUDA

Question

How do I allocate and transfer(to and from Host) 2D arrays in device memory in Cuda?

Answer 1

Flatten it: make it one-dimensional. See how it's done here

Answer 2

I found a solution to this problem. I didn't have to flatten the array.

The inbuild cudaMallocPitch() function did the job. And I could transfer the array to and from device using cudaMemcpy2D() function.

for eg,

cudaMallocPitch((void**) &array, &pitch, a*sizeof(float), b);

This creates a 2D array of size a*b with the pitch as passed in as parameter.

The following code creates a 2D array and loops over the elements. It compiles readily, you may use it.

#include<stdio.h>
#include<cuda.h>
#define height 50
#define width 50

// Device code
__global__ void kernel(float* devPtr, int pitch)
{
    for (int r = 0; r < height; ++r) {
        float* row = (float*)((char*)devPtr + r * pitch);
        for (int c = 0; c < width; ++c) {
             float element = row[c];
        }
    }
}

//Host Code
int main()
{

float* devPtr;
size_t pitch;
cudaMallocPitch((void**)&devPtr, &pitch, width * sizeof(float), height);
kernel<<<100, 512>>>(devPtr, pitch);
return 0;
}

Answer 3

How did you declare the parameters in the function as int **d_A where d_A[][] is the 2-D device array that we need? When i do that it doesnt clearly know the dimensions and gives some wierd answers. I too used cudaMallocPitch and cudaMalloc2D but this was causing me trouble.

Please advise