使用 cudamalloc().为什么是双指针? [英] Use of cudamalloc(). Why the double pointer?

问题描述

我目前正在浏览 http://code.google.com/p/stanford 上的教程示例-cs193g-sp2010/ 来学习 CUDA.下面给出了演示 __global__ 函数的代码.它只是创建了两个数组，一个在 CPU 上，一个在 GPU 上，用数字 7 填充 GPU 数组并将 GPU 数组数据复制到 CPU 数组中.

I am currently going through the tutorial examples on http://code.google.com/p/stanford-cs193g-sp2010/ to learn CUDA. The code which demostrates __global__ functions is given below. It simply creates two arrays, one on the CPU and one on the GPU, populates the GPU array with the number 7 and copies the GPU array data into the CPU array.

#include <stdlib.h>
#include <stdio.h>

__global__ void kernel(int *array)
{
  int index = blockIdx.x * blockDim.x + threadIdx.x;

  array[index] = 7;
}

int main(void)
{
  int num_elements = 256;

  int num_bytes = num_elements * sizeof(int);

  // pointers to host & device arrays
  int *device_array = 0;
  int *host_array = 0;

  // malloc a host array
  host_array = (int*)malloc(num_bytes);

  // cudaMalloc a device array
  cudaMalloc((void**)&device_array, num_bytes);

  int block_size = 128;
  int grid_size = num_elements / block_size;

  kernel<<<grid_size,block_size>>>(device_array);

  // download and inspect the result on the host:
  cudaMemcpy(host_array, device_array, num_bytes, cudaMemcpyDeviceToHost);

  // print out the result element by element
  for(int i=0; i < num_elements; ++i)
  {
    printf("%d ", host_array[i]);
  }

  // deallocate memory
  free(host_array);
  cudaFree(device_array);
} 

我的问题是为什么他们用双指针来编写 cudaMalloc((void**)&device_array, num_bytes); 语句?甚至 here 定义的 cudamalloc() on 说第一个参数是双指针.

My question is why have they worded the cudaMalloc((void**)&device_array, num_bytes); statement with a double pointer? Even here definition of cudamalloc() on says the first argument is a double pointer.

为什么不像 CPU 上的 malloc 函数那样简单地返回一个指向 GPU 上分配内存开头的指针?

Why not simply return a pointer to the beginning of the allocated memory on the GPU, just like the malloc function does on the CPU?

推荐答案

所有 CUDA API 函数都返回一个错误代码(如果没有发生错误，则返回 cudaSuccess).所有其他参数都通过引用传递.但是，在纯 C 中，您不能有引用，这就是为什么您必须传递您希望存储返回信息的变量的地址.由于要返回指针，因此需要传递双指针.

All CUDA API functions return an error code (or cudaSuccess if no error occured). All other parameters are passed by reference. However, in plain C you cannot have references, that's why you have to pass an address of the variable that you want the return information to be stored. Since you are returning a pointer, you need to pass a double-pointer.

另一个众所周知的基于同样原因对地址进行操作的函数是 scanf 函数.您有多少次忘记在要存储值的变量之前写此 & ?;)

Another well-known function which operates on addresses for the same reason is the scanf function. How many times have you forgotten to write this & before the variable that you want to store the value to? ;)

int i;
scanf("%d",&i);

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