std::vector to array in CUDA

Question

Is there a way to convert a 2D vector into an array to be able to use it in CUDA kernels?

It is declared as:

vector<vector<int>> information;

I want to cudaMalloc and copy from host to device, what would be the best way to do it?

int *d_information;
cudaMalloc((void**)&d_information, sizeof(int)*size);
cudaMemcpy(d_information, information, sizeof(int)*size, cudaMemcpyHostToDevice);

Davide Spataro · Accepted Answer · 2021-01-09 15:28:02Z

4

In a word, no there isn't. The CUDA API doesn't support deep copying and also doesn't know anything about std::vector either. If you insist on having a vector of vectors as a host source, it will require doing something like this:

int *d_information;
cudaMalloc((void**)&d_information, sizeof(int)*size);

int *dst = d_information;
for (std::vector<std::vector<int> >::iterator it = information.begin() ; it != information.end(); ++it) {
    int *src = &((*it)[0]);
    size_t sz = it->size();
    
    cudaMemcpy(dst, src, sizeof(int)*sz, cudaMemcpyHostToDevice);
    dst += sz;
}

[disclaimer: written in browser, not compiled or tested. Use at own risk]

This would copy the host memory to an allocation in GPU linear memory, requiring one copy for each vector. If the vector of vectors is a "jagged" array, you will want to store an indexing somewhere for the GPU to use as well.

edited Jan 9, 2021 at 15:28

Davide Spataro

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answered Jul 10, 2013 at 12:33

talonmies

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4 Comments

BRabbit27 Over a year ago

Ok, so no way having std::vector in CUDA. Could it be helpful (and easy) to use Thrust? I've never used it before but as far as I read it's somehow similar to the STL library but in CUDA. Any advise?

talonmies Over a year ago

No, thrust doesn't have any support for this either. You would be much better off just flattening your host array into a std::vector<int> and indexing it as you would linear memory on the device.

talonmies Over a year ago

@BRabbit27: I don't know why your edit was rejected, it was correct. I dashed off that code in the browser and when I put that disclaimer in, I really mean it.

BRabbit27 Over a year ago

yeah I analyzed what you proposed, I found the mistake and try it on my code and well it worked. I knew the disclaimer was there because you meant it. Anyways, I corrected it use in case someone is interested in something like that but still the disclaimer should be taken into account.

perror · Accepted Answer · 2013-07-10 12:53:04Z

2

As far as I understand, the vector of vectors do not need to reside in a contiguous memory, i.e. they can be fragmented.

Depending on the amount of memory you need to transfer I would do one of two issues:

Reorder your memory to be a single vector, and then use your cudaMemcpy.
Create a series of cudaMemcpyAsync, where each copy handles a single vector in your vector of vectors, and then synchronize.

edited Jul 10, 2013 at 12:53

perror

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answered Jul 10, 2013 at 12:29

Henrik

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