如何从 CUDA 内核函数返回单个变量？

发布于 2024-08-29 01:03:00 字数 336 浏览 2 评论 0原文

我有一个 CUDA 搜索函数，可以计算一个变量。我怎样才能把它退回来。

__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);

对于这两行我都收到此错误：错误：“long”类型的参数与“const void *”类型的参数不兼容

原文

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 *"

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情栀口红 2024-09-05 01:03:00

为此，我一直使用 __device__ 变量，这样您就不必担心 cudaMalloc 和 cudaFree必须传递一个指针作为内核参数，这会在内核中保存一个寄存器来启动。

__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;
}

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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月亮坠入山谷 2024-09-05 01:03:00

要获得单个结果，您必须对其进行 Memcpy，即：

#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;
}

我猜错误的出现是因为您传递的是长值，而不是指向长值的指针。

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.

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