Passing a 2D Array to CUDA Device and Using it [duplicate]

Question

I have a problem related to CUDA. I create a array which is 2D with :

char **WordMatrix = new char*[N]; // M and N set by the user
for(int i = 0; i < N; ++i) 
{
    WordMatrix[i] = new char[M];
}

I fill the array with a loop. After that where my problem is lying : I can't managed to allocate the memory in device and copy from host to device.

I saw other questions related to this but I did not understand the logic. Like in this topic : https://devtalk.nvidia.com/default/topic/410182/double-pointer-allocation/ or in here.

I want to understand how to do it with an code example which uses cudaMemcpy and cudaMalloc and 2D Array also explanation of why we need supporting(if you use the approach in the link) pointer?

Community · Accepted Answer · 2017-05-23 12:10:04Z

4

My understanding is CUDA accepts only linearized 2D arrays, so really a 1D array.

int  *my_array = new int[height*width];
for (int h = 0; h < height; h++){
    for (int w = 0; w < width; w++)
        my_array[width * h + w] = value;
}

You can then copy that to device memory as in the other answer.

Also, this question has more info: Allocate 2D Array on Device Memory in CUDA.

edited May 23, 2017 at 12:10

CommunityBot

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answered Jul 12, 2014 at 17:08

Gswffye

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1 Comment

Robert Crovella Over a year ago

"CUDA accepts only linearized 2D arrays" This is not correct. CUDA can work with 2D arrays. The marked duplicate question demonstrates this. However it's usually not recommended.

dari · Accepted Answer · 2014-07-12 16:44:12Z

2

Create your 2D array in one piece new char[N*M]
Allocate the same amout of memory on GPU cudaMalloc(... sizeof(char)*N*M)
Copy your CPU memory (1.) to GPU (2.) with cudaMemcpy(... hostToDevice)

answered Jul 12, 2014 at 16:44

dari

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1 Comment

Oguzhan D. Over a year ago

Isn't this kind of creating 1D array?

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