get reverse_iterator from device_ptr in CUDA - cuda

Question

For a device_vector, I can use its rbegin() method to get its reverse iterator. But how to construct a reverse iterator directly from device_ptr?
May be this can be achieved by constructing a device_vector with the device_ptr, the code is as follows:
thrust::device_ptr<int> ptr = get_ptr();
thrust::device_vector<int> tmpVector(ptr , ptr + N)
thrust::inclusive_scan_by_key(tmpVector.rbegin(), tmpVector.rend(), ......);
But I don't know if thrust::device_vector<int> tmpVector(ptr , ptr + N) will construct a new vector and copy the data from ptr or it just reserve a reference from ptr? The documentation of Thrust doesn't mension this.
Any ideas?

Answer 1

Providing an answer based on the comment by Jared, to get this off the unanswered list, and to preserve the question for future readers.
To make a reverse iterator from any kind of iterator, including thrust::device_ptr, use the thrust::make_reverse_iterator function.
Here is a simple example:
$ cat t615.cu
#include <thrust/device_vector.h>
#include <thrust/iterator/reverse_iterator.h>
#include <thrust/device_ptr.h>
#include <thrust/sequence.h>
#include <thrust/copy.h>
#include <iostream>
#define DSIZE 4
int main(){
int *data;
cudaMalloc(&data, DSIZE*sizeof(int));
thrust::device_ptr<int> my_data = thrust::device_pointer_cast<int>(data);
thrust::sequence(my_data, my_data+DSIZE);
thrust::copy_n(my_data, DSIZE, std::ostream_iterator<int>(std::cout, ","));
std::cout << std::endl;
typedef thrust::device_vector<int>::iterator Iterator;
thrust::reverse_iterator<Iterator> r_iter = make_reverse_iterator(my_data+DSIZE); // note that we point the iterator to the "end" of the device pointer area
thrust::copy_n(r_iter, DSIZE, std::ostream_iterator<int>(std::cout, ","));
std::cout << std::endl;
return 0;
}
$ nvcc -arch=sm_35 -o t615 t615.cu
$ ./t615
0,1,2,3,
3,2,1,0,
$
The creation of a reverse iterator does not create any "extra array".