CUDA thrust: copy from device to device

2018-06-16 17:48:47

I have a memory array allocated in CUDA using standard CUDA malloc and it is passed to a function as follows:

void MyClass::run(uchar4 * input_data)

I also have a class member which is a thrust device_ptr declared as:

thrust::device_ptr<uchar4> data = thrust::device_malloc<uchar4(num_pts);

Here num_pts is the number of values in the array and the input_data pointer is guaranteed to be num_pts long.

Now, I would like to copy the input array into the thrust_device_ptr. I have looked at the thrust documentation and a lot of it is talking about copying from device to host memory and vice versa. I was wondering what would be the most performance optimal way to do this device to device copy on thrust or should I just use cudaMemcpy?

The canonical way to do this is just to use thrust::copy . The thrust::device_ptr has standard pointer semantics and the API will seamlessly understand whether the source and destination pointers are on the host or device, viz:

#include <thrust/device_malloc.h>
#include <thrust/device_ptr.h>
#include <thrust/copy.h>
#include <iostream>

int main()
{
    // Initial host data
    int ivals[4] = { 1, 3, 6, 10 };

    // Allocate and copy to first device allocation
    thrust::device_ptr<int> dp1 = thrust::device_malloc<int>(4);
    thrust::copy(&ivals[0], &ivals[0]+4, dp1);

    // Allocate and copy to second device allocation
    thrust::device_ptr<int> dp2 = thrust::device_malloc<int>(4);
    thrust::copy(dp1, dp1+4, dp2);

    // Copy back to host
    int ovals[4] = {-1, -1, -1, -1};
    thrust::copy(dp2, dp2+4, &ovals[0]);

    for(int i=0; i<4; i++)
        std::cout << ovals[i] << std::endl;


    return 0;
}

which does this:

talonmies@box:~$ nvcc -arch=sm_30 thrust_dtod.cu 
talonmies@box:~$ ./a.out 
1
3
6
10

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