Compute reduction sum of a device array with thrust

Question

I know we can compute sum of a CPU(host) array with thrust like this.

int data[6] = {1, 0, 2, 2, 1, 3};
int result = thrust::reduce(data, data + 6, 0);

Can we find sum of GPU array with thrust without cudaMemcpy to CPU array?
Suppose I have a device array created using cudaMalloc like this,

cudaMalloc(&gpuspeed, n* sizeof(int));

and did modifications to gpuspeed with some kernels. Now can I find sum of that with thrust? If we can, what changes I have to make?

score 1 · Accepted Answer · answered May 07 '21 at 16:50

Yes, you can do that with thrust.

You can pass device pointers to thrust, and thrust will do the right thing if you specify explicitly the device execution path, using thrust execution policies.

Alternatively, you can use thrust::device_ptr to refer to your data, and thrust will also do the right thing, even without explicitly specifying the device execution path.

This answer covers both approaches, albeit with inclusive_scan.

Here is an example:

$ cat t137.cu
#include <thrust/reduce.h>
#include <thrust/device_ptr.h>
#include <thrust/execution_policy.h>
#include <iostream>

__global__ void k(int *d, int n){
  int idx = threadIdx.x+blockDim.x*blockIdx.x;
  if (idx < n)
    d[idx] = idx;
}
const int ds = 10;
const int nTPB = 256;
int main(){

  int *d, r1, r2;
  cudaMalloc(&d, ds*sizeof(d[0]));
  k<<<(ds+nTPB-1)/nTPB,nTPB>>>(d, ds);
  thrust::device_ptr<int> tdp = thrust::device_pointer_cast(d);
  r1 = thrust::reduce(tdp, tdp+ds);
  r2 = thrust::reduce(thrust::device, d, d+ds);
  std::cout << "r1: "  << r1 << " r2: " << r2 << std::endl;
}
$ nvcc -std=c++14 -o t137 t137.cu
$ ./t137
r1: 45 r2: 45
$

Compute reduction sum of a device array with thrust

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