How to return a single variable from a CUDA kernel function?

21
votes

I have a CUDA search function which calculate one single variable. How can I return it back.

__global__ 
void G_SearchByNameID(node* Node, long nodeCount, long start,char* dest, long answer){
    answer = 2;
}

cudaMemcpy(h_answer, d_answer, sizeof(long), cudaMemcpyDeviceToHost);
cudaFree(d_answer);

for both of these lines I get this error: error: argument of type "long" is incompatible with parameter of type "const void *"

2
ccuda

2 Answers

33
votes

I've been using __device__ variables for this purpose, that way you don't have to bother with cudaMalloc and cudaFree and you don't have to pass a pointer as a kernel argument, which saves you a register in your kernel to boot.

__device__ long d_answer;

__global__ void G_SearchByNameID() {
  d_answer = 2;
}

int main() {
  SearchByNameID<<<1,1>>>();
  typeof(d_answer) answer;
  cudaMemcpyFromSymbol(&answer, "d_answer", sizeof(answer), 0, cudaMemcpyDeviceToHost);
  printf("answer: %d\n", answer);
  return 0;
}
23
votes

To get a single result you have to Memcpy it, ie:

#include <assert.h>

__global__ void g_singleAnswer(long* answer){ *answer = 2; }

int main(){

  long h_answer;
  long* d_answer;
  cudaMalloc(&d_answer, sizeof(long));
  g_singleAnswer<<<1,1>>>(d_answer);
  cudaMemcpy(&h_answer, d_answer, sizeof(long), cudaMemcpyDeviceToHost); 
  cudaFree(d_answer);
  assert(h_answer == 2);
  return 0;
}

I guess the error come because you are passing a long value, instead of a pointer to a long value.