如何从CUDA内核函数返回单个变量? [英] How to return a single variable from a CUDA kernel function?

问题描述

我有一个CUDA搜索功能，可以计算一个单个变量.如何退回.

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);

对于这两行，我都会收到此错误:错误:"long"类型的参数与"const void *"类型的参数不兼容

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

推荐答案

为此，我一直在使用 __ device __ 变量，这样您就不必为 cudaMalloc 和 cudaFree ，您不必将指针作为内核参数传递，这样可以节省内核中的寄存器以进行引导.

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;
}

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