cudaMallocHost vs malloc表现更好,没有任何区别

Question

我已经浏览了这个网站.从这里我得到了使用cudamallocHost的固定内存提供了比cudamalloc更好的性能.然后我使用两个不同的简单程序并测试执行时间为

使用cudaMallocHost

#include <stdio.h>
#include <cuda.h>

// Kernel that executes on the CUDA device
__global__ void square_array(float *a, int N)
{
  int idx = blockIdx.x * blockDim.x + threadIdx.x;
  if (idx<N) a[idx] = a[idx] * a[idx];
}

// main routine that executes on the host
int main(void)
{
    clock_t start;
    start=clock();/* Line 8 */
    clock_t finish;
  float *a_h, *a_d;  // Pointer to host & device arrays
  const int N = 100000;  // Number of elements in arrays
  size_t size = N * sizeof(float);
  cudaMallocHost((void **) &a_h, size);
  //a_h = (float *)malloc(size);        // Allocate array on host
  cudaMalloc((void **) &a_d, size);   // Allocate array on device
  // Initialize host array and copy it to CUDA device
  for (int i=0; i<N; i++) a_h[i] = (float)i;
  cudaMemcpy(a_d, a_h, size, cudaMemcpyHostToDevice);
  // Do calculation on device:
  int block_size = 4;
  int n_blocks = N/block_size + (N%block_size == 0 ? 0:1);
  square_array <<< n_blocks, block_size >>> (a_d, N);
  // Retrieve result from device and store it in host array
  cudaMemcpy(a_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost);
  // Print results
  for (int i=0; i<N; i++) printf("%d %f\n", i, a_h[i]);
  // Cleanup
  cudaFreeHost(a_h);
  cudaFree(a_d);
  finish = clock() - start;
      double interval = finish / (double)CLOCKS_PER_SEC; 
      printf("%f seconds elapsed", interval);
}

使用Malloc

#include <stdio.h>
#include <cuda.h>

// Kernel that executes on the CUDA device
__global__ void square_array(float *a, int N)
{
  int idx = blockIdx.x * blockDim.x + threadIdx.x;
  if (idx<N) a[idx] = a[idx] * a[idx];
}

// main routine that executes on the host
int main(void)
{
    clock_t start;
    start=clock();/* Line 8 */
    clock_t finish;
  float *a_h, *a_d;  // Pointer to host & device arrays
  const int N = 100000;  // Number of elements in arrays
  size_t size = N * sizeof(float);
  a_h = (float *)malloc(size);        // Allocate array on host
  cudaMalloc((void **) &a_d, size);   // Allocate array on device
  // Initialize host array and copy it to CUDA device
  for (int i=0; i<N; i++) a_h[i] = (float)i;
  cudaMemcpy(a_d, a_h, size, cudaMemcpyHostToDevice);
  // Do calculation on device:
  int block_size = 4;
  int n_blocks = N/block_size + (N%block_size == 0 ? 0:1);
  square_array <<< n_blocks, block_size >>> (a_d, N);
  // Retrieve result from device and store it in host array
  cudaMemcpy(a_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost);
  // Print results
  for (int i=0; i<N; i++) printf("%d %f\n", i, a_h[i]);
  // Cleanup
  free(a_h); cudaFree(a_d);
  finish = clock() - start;
      double interval = finish / (double)CLOCKS_PER_SEC; 
      printf("%f seconds elapsed", interval);
}

在执行这两个程序时,执行时间几乎相似.实施中有什么问题吗？在cudamalloc和cudamallochost执行的确切区别是什么？

并且每次运行时执行时间减少

Answer 1

如果要查看复制操作的执行时间差异,只需复制操作即可.在许多情况下,当基础记忆固定时,您只会看到复制操作的执行时间差异大约为2 倍.并使您的复制操作足够大/足够长,以便您远远超出您正在使用的任何计时机制的粒度.各种分析器,如视觉分析器,nvprof可以在这里提供帮助.

在cudaMallocHost引擎盖下的操作是做一些像malloc再加上额外的OS功能为"针"与分配相关联的每一页.与仅执行操作相比,这些额外的操作系统操作需要额外的时间malloc.请注意,随着分配规模的增加,注册("钉扎")成本通常也会增加.

因此,对于许多示例,仅仅对整体执行的计时没有显示出太大的差异,因为虽然cudaMemcpy操作可能cudaMallocHost比固定存储器更快,但是花费的时间比相应的长malloc.

那有什么意义呢？

1. cudaMallocHost当您从单个缓冲区重复传输时,您可能会对使用固定内存(即)感兴趣.您只需支付额外费用即可固定一次,但每次转帐/使用都会受益.
2. 固定内存需要将数据传输操作(cudaMemcpyAsync)与计算活动(内核调用)重叠.请参阅编程指南.