当使用cub :: BlockRadixSort在块内进行排序时,如果元素的数量太大,我们该如何处理?如果我们将磁贴大小设置得太大,临时存储的共享内存很快就无法保存。如果我们将它拆分为多个图块,我们如何在对每个图块进行排序后对其进行后处理?
答案 0 :(得分:3)
根据我的测试,cub并不真正了解原始数据的位置或放置临时存储的位置。因此,一种可能的解决方案就是将临时存储放在全局内存中。为了分析这个,我创建了一个包含3个不同测试用例的代码:
这些都没有经过广泛的测试,但由于我正在构建Cub构建块,并在前两种情况下测试我的结果,所以希望我没有犯过任何严重的错误。这是完整的测试代码,我将在下面做出更多评论:
$ cat t10.cu
#include <cub/cub.cuh>
#include <stdio.h>
#include <stdlib.h>
#include <thrust/sort.h>
#define nTPB 512
#define ELEMS_PER_THREAD 2
#define RANGE (nTPB*ELEMS_PER_THREAD)
#define DSIZE (nTPB*ELEMS_PER_THREAD)
#define cudaCheckErrors(msg) \
do { \
cudaError_t __err = cudaGetLastError(); \
if (__err != cudaSuccess) { \
fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
msg, cudaGetErrorString(__err), \
__FILE__, __LINE__); \
fprintf(stderr, "*** FAILED - ABORTING\n"); \
exit(1); \
} \
} while (0)
using namespace cub;
// GLOBAL CUB BLOCK SORT KERNEL
// Specialize BlockRadixSort collective types
typedef BlockRadixSort<int, nTPB, ELEMS_PER_THREAD> my_block_sort;
__device__ int my_val[DSIZE];
__device__ typename my_block_sort::TempStorage sort_temp_stg;
// Block-sorting CUDA kernel (nTPB threads each owning ELEMS_PER THREAD integers)
__global__ void global_BlockSortKernel()
{
// Collectively sort the keys
my_block_sort(sort_temp_stg).Sort(*static_cast<int(*)[ELEMS_PER_THREAD]>(static_cast<void*>(my_val+(threadIdx.x*ELEMS_PER_THREAD))));
}
// ORIGINAL CUB BLOCK SORT KERNEL
template <int BLOCK_THREADS, int ITEMS_PER_THREAD>
__global__ void BlockSortKernel(int *d_in, int *d_out)
{
// Specialize BlockLoad, BlockStore, and BlockRadixSort collective types
typedef cub::BlockLoad<int*, BLOCK_THREADS, ITEMS_PER_THREAD, BLOCK_LOAD_TRANSPOSE> BlockLoadT;
typedef cub::BlockStore<int*, BLOCK_THREADS, ITEMS_PER_THREAD, BLOCK_STORE_TRANSPOSE> BlockStoreT;
typedef cub::BlockRadixSort<int, BLOCK_THREADS, ITEMS_PER_THREAD> BlockRadixSortT;
// Allocate type-safe, repurposable shared memory for collectives
__shared__ union {
typename BlockLoadT::TempStorage load;
typename BlockStoreT::TempStorage store;
typename BlockRadixSortT::TempStorage sort;
} temp_storage;
// Obtain this block's segment of consecutive keys (blocked across threads)
int thread_keys[ITEMS_PER_THREAD];
int block_offset = blockIdx.x * (BLOCK_THREADS * ITEMS_PER_THREAD);
BlockLoadT(temp_storage.load).Load(d_in + block_offset, thread_keys);
__syncthreads(); // Barrier for smem reuse
// Collectively sort the keys
BlockRadixSortT(temp_storage.sort).Sort(thread_keys);
__syncthreads(); // Barrier for smem reuse
// Store the sorted segment
BlockStoreT(temp_storage.store).Store(d_out + block_offset, thread_keys);
}
// SHARED MEM CUB BLOCK SORT KERNEL
// Block-sorting CUDA kernel (nTPB threads each owning ELEMS_PER THREAD integers)
template <int BLOCK_THREADS, int ITEMS_PER_THREAD>
__global__ void shared_BlockSortKernel(int *d_out)
{
__shared__ int my_val[BLOCK_THREADS*ITEMS_PER_THREAD];
// Specialize BlockRadixSort collective types
typedef BlockRadixSort<int, BLOCK_THREADS, ITEMS_PER_THREAD> my_block_sort;
// Allocate shared memory for collectives
__shared__ typename my_block_sort::TempStorage sort_temp_stg;
// need to extend synthetic data for ELEMS_PER_THREAD > 1
my_val[threadIdx.x*ITEMS_PER_THREAD] = (threadIdx.x + 5); // synth data
my_val[threadIdx.x*ITEMS_PER_THREAD+1] = (threadIdx.x + BLOCK_THREADS + 5); // synth data
__syncthreads();
// printf("thread %d data = %d\n", threadIdx.x, my_val[threadIdx.x*ITEMS_PER_THREAD]);
// Collectively sort the keys
my_block_sort(sort_temp_stg).Sort(*static_cast<int(*)[ITEMS_PER_THREAD]>(static_cast<void*>(my_val+(threadIdx.x*ITEMS_PER_THREAD))));
__syncthreads();
// printf("thread %d sorted data = %d\n", threadIdx.x, my_val[threadIdx.x*ITEMS_PER_THREAD]);
if (threadIdx.x == clock()){ // dummy to prevent compiler optimization
d_out[threadIdx.x*ITEMS_PER_THREAD] = my_val[threadIdx.x*ITEMS_PER_THREAD];
d_out[threadIdx.x*ITEMS_PER_THREAD+1] = my_val[threadIdx.x*ITEMS_PER_THREAD+1];}
}
int main(){
int *h_data, *h_result;
cudaEvent_t start, stop;
cudaEventCreate(&start);
cudaEventCreate(&stop);
h_data=(int *)malloc(DSIZE*sizeof(int));
h_result=(int *)malloc(DSIZE*sizeof(int));
if (h_data == 0) {printf("malloc fail\n"); return 1;}
if (h_result == 0) {printf("malloc fail\n"); return 1;}
for (int i = 0 ; i < DSIZE; i++) h_data[i] = rand()%RANGE;
// first test sorting directly out of global memory
global_BlockSortKernel<<<1,nTPB>>>(); //warm up run
cudaDeviceSynchronize();
cudaMemcpyToSymbol(my_val, h_data, DSIZE*sizeof(int));
cudaCheckErrors("memcpy to symbol fail");
cudaEventRecord(start);
global_BlockSortKernel<<<1,nTPB>>>(); //timing run
cudaEventRecord(stop);
cudaDeviceSynchronize();
cudaCheckErrors("cub 1 fail");
cudaEventSynchronize(stop);
float et;
cudaEventElapsedTime(&et, start, stop);
cudaMemcpyFromSymbol(h_result, my_val, DSIZE*sizeof(int));
cudaCheckErrors("memcpy from symbol fail");
if(!thrust::is_sorted(h_result, h_result+DSIZE)) { printf("sort 1 fail!\n"); return 1;}
printf("global Elapsed time: %fms\n", et);
printf("global Kkeys/s: %d\n", (int)(DSIZE/et));
// now test original CUB block sort copying global to shared
int *d_in, *d_out;
cudaMalloc((void **)&d_in, DSIZE*sizeof(int));
cudaMalloc((void **)&d_out, DSIZE*sizeof(int));
cudaCheckErrors("cudaMalloc fail");
BlockSortKernel<nTPB, ELEMS_PER_THREAD><<<1, nTPB>>>(d_in, d_out); // warm up run
cudaMemcpy(d_in, h_data, DSIZE*sizeof(int), cudaMemcpyHostToDevice);
cudaEventRecord(start);
BlockSortKernel<nTPB, ELEMS_PER_THREAD><<<1, nTPB>>>(d_in, d_out); // timing run
cudaEventRecord(stop);
cudaDeviceSynchronize();
cudaCheckErrors("cub 2 fail");
cudaEventSynchronize(stop);
cudaEventElapsedTime(&et, start, stop);
cudaMemcpy(h_result, d_out, DSIZE*sizeof(int), cudaMemcpyDeviceToHost);
cudaCheckErrors("cudaMemcpy D to H fail");
if(!thrust::is_sorted(h_result, h_result+DSIZE)) { printf("sort 2 fail!\n"); return 1;}
printf("CUB Elapsed time: %fms\n", et);
printf("CUB Kkeys/s: %d\n", (int)(DSIZE/et));
// now test shared memory-only version of block sort
shared_BlockSortKernel<nTPB, ELEMS_PER_THREAD><<<1, nTPB>>>(d_out); // warm-up run
cudaEventRecord(start);
shared_BlockSortKernel<nTPB, ELEMS_PER_THREAD><<<1, nTPB>>>(d_out); // timing run
cudaEventRecord(stop);
cudaDeviceSynchronize();
cudaCheckErrors("cub 3 fail");
cudaEventSynchronize(stop);
cudaEventElapsedTime(&et, start, stop);
printf("shared Elapsed time: %fms\n", et);
printf("shared Kkeys/s: %d\n", (int)(DSIZE/et));
return 0;
}
$ nvcc -O3 -arch=sm_20 -o t10 t10.cu
$ ./t10
global Elapsed time: 0.236960ms
global Kkeys/s: 4321
CUB Elapsed time: 0.042816ms
CUB Kkeys/s: 23916
shared Elapsed time: 0.040192ms
shared Kkeys/s: 25477
$
对于这个测试,我使用的是CUDA 6.0RC,cub v1.2.0(这是最近的版本),RHEL5.5 / gcc4.1.2和Quadro5000 GPU(cc2.0,11SMs,比a慢了约40%) GTX480)。以下是我发现的一些观察结果:
6:1的罚款是一个很大的罚款。所以我的建议是,如果可能的话,对问题大小使用设备范围的排序,大于可以通过Cub块排序轻松处理的问题。这使您可以利用一些最佳GPU代码编写器的专业知识进行排序,并实现更接近整个设备能力的吞吐量。
请注意,我可以在类似条件下进行测试,此处的问题大小(512个线程,每个线程2个元素)不会超过您在CUB块排序中可以执行的操作。但是,将数据集大小扩展为更大的值(例如,每个线程1024个元素)并不难,只能使用第一种方法处理(在这种情况下,在这些选择中)。如果我做了更大的问题大小,在我的GPU上,我观察到我的cc2.0设备上的全局内存块排序的吞吐量约为6Mkeys / s。