Thrust array of device_vectors to raw_pointer

Question

Using device_vector:

thrust::device_vector< int > iVec;
int* iArray = thrust::raw_pointer_cast( &iVec[0] );

but how can I do it if I have an array of device_vectors?

thrust::device_vector<int> iVec[10];

Ideally I would like to pass my array of device_vector to a 1D array to be handled on a CUDA kernel. Is it possible?

You understand that a device_vector is a C++ class in host memory? An array of device_vector is an array of C++ class instances, also in host memory. There is no raw device pointer because there is no device memory involved. — talonmies
– talonmies, Commented Oct 17, 2012 at 11:24
@talonmies: So,there is no way I can covert the array of device_vector to 1D host array? — Manolete
– Manolete, Commented Oct 17, 2012 at 11:36
An array of device vectors is already 1D host array. What is it that you are really trying to do? — talonmies
– talonmies, Commented Oct 17, 2012 at 11:47
I want to pass the array to a kernel so I can deal with it with all the elements in order, thinking that every array might have different length of the vector. — Manolete
– Manolete, Commented Oct 17, 2012 at 11:53

talonmies · Accepted Answer · 2012-10-18 06:14:06Z

1

If I have understood your question correctly, what you are really try to do is create an array of raw pointers from an array of thrust::device_vectors. You should be able to do this like so:

const int N = 10;
thrust::device_vector<int> iVec[N];

int * iRaw[N];
for(int i=0; i<N; i++)
    iRaw[i] = thrust::raw_pointer_cast(iVec[i].data());

int ** _iRaw;
size_t sz = sizeof(int *) * N;
cudaMalloc((void ***)&_iRaw, sz);
cudaMemcpy(_iRaw, iRaw, sz, cudaMemcpyHostToDevice);

[disclaimer: written in browser, never compiled, never tested, use at own risk]

In the above code snippet, _iRaw holds the raw pointers of each of the device vectors in iVec. You could pass that to a kernel if you really wanted to.

edited Oct 18, 2012 at 6:14

answered Oct 17, 2012 at 16:33

talonmies

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