让CUDA推动使用您选择的CUDA流

Question

看看CUDA Thrust代码中的内核启动,它们似乎总是使用默认流.我可以让Thrust使用我选择的流吗？我错过了API中的内容吗？

Answer 1

我想在Thrust 1.8发布后更新talonmies提供的答案,它引入了将CUDA执行流指示为

thrust::cuda::par.on(stream)

也可以看看

推力释放1.8.0.

在下面,我正在重述这个例子

Fermi架构的错误依赖问题

就CUDA Thrust API而言.

#include <iostream>

#include "cuda_runtime.h"
#include "device_launch_parameters.h"

#include <stdio.h>

#include <thrust\device_vector.h>
#include <thrust\execution_policy.h>

#include "Utilities.cuh"

using namespace std;

#define NUM_THREADS 32
#define NUM_BLOCKS 16
#define NUM_STREAMS 3

struct BinaryOp{ __host__ __device__ int operator()(const int& o1,const int& o2) { return o1 * o2; } };

int main()
{
    const int N = 6000000;

    // --- Host side input data allocation and initialization. Registering host memory as page-locked (required for asynch cudaMemcpyAsync).
    int *h_in = new int[N]; for(int i = 0; i < N; i++) h_in[i] = 5;
    gpuErrchk(cudaHostRegister(h_in, N * sizeof(int), cudaHostRegisterPortable));

    // --- Host side input data allocation and initialization. Registering host memory as page-locked (required for asynch cudaMemcpyAsync).
    int *h_out = new int[N]; for(int i = 0; i < N; i++) h_out[i] = 0;
    gpuErrchk(cudaHostRegister(h_out, N * sizeof(int), cudaHostRegisterPortable));

    // --- Host side check results vector allocation and initialization
    int *h_checkResults = new int[N]; for(int i = 0; i < N; i++) h_checkResults[i] = h_in[i] * h_in[i];

    // --- Device side input data allocation.
    int *d_in = 0;              gpuErrchk(cudaMalloc((void **)&d_in, N * sizeof(int)));

    // --- Device side output data allocation. 
    int *d_out = 0;             gpuErrchk( cudaMalloc((void **)&d_out, N * sizeof(int)));

    int streamSize = N / NUM_STREAMS;
    size_t streamMemSize = N * sizeof(int) / NUM_STREAMS;

    // --- Set kernel launch configuration
    dim3 nThreads       = dim3(NUM_THREADS,1,1);
    dim3 nBlocks        = dim3(NUM_BLOCKS, 1,1);
    dim3 subKernelBlock = dim3((int)ceil((float)nBlocks.x / 2));

    // --- Create CUDA streams
    cudaStream_t streams[NUM_STREAMS];
    for(int i = 0; i < NUM_STREAMS; i++)
        gpuErrchk(cudaStreamCreate(&streams[i]));

    /**************************/
    /* BREADTH-FIRST APPROACH */
    /**************************/

    for(int i = 0; i < NUM_STREAMS; i++) {
        int offset = i * streamSize;
        cudaMemcpyAsync(&d_in[offset], &h_in[offset], streamMemSize, cudaMemcpyHostToDevice,     streams[i]);
    }

    for(int i = 0; i < NUM_STREAMS; i++)
    {
        int offset = i * streamSize;

        thrust::transform(thrust::cuda::par.on(streams[i]), thrust::device_pointer_cast(&d_in[offset]), thrust::device_pointer_cast(&d_in[offset]) + streamSize/2, 
                                                            thrust::device_pointer_cast(&d_in[offset]), thrust::device_pointer_cast(&d_out[offset]), BinaryOp());
        thrust::transform(thrust::cuda::par.on(streams[i]), thrust::device_pointer_cast(&d_in[offset + streamSize/2]), thrust::device_pointer_cast(&d_in[offset + streamSize/2]) + streamSize/2, 
                                                            thrust::device_pointer_cast(&d_in[offset + streamSize/2]), thrust::device_pointer_cast(&d_out[offset + streamSize/2]), BinaryOp());

    }

    for(int i = 0; i < NUM_STREAMS; i++) {
        int offset = i * streamSize;
        cudaMemcpyAsync(&h_out[offset], &d_out[offset], streamMemSize, cudaMemcpyDeviceToHost,   streams[i]);
    }

    for(int i = 0; i < NUM_STREAMS; i++)
        gpuErrchk(cudaStreamSynchronize(streams[i]));

    gpuErrchk(cudaDeviceSynchronize());

    // --- Release resources
    gpuErrchk(cudaHostUnregister(h_in));
    gpuErrchk(cudaHostUnregister(h_out));
    gpuErrchk(cudaFree(d_in));
    gpuErrchk(cudaFree(d_out));

    for(int i = 0; i < NUM_STREAMS; i++)
        gpuErrchk(cudaStreamDestroy(streams[i]));

    cudaDeviceReset();  

    // --- GPU output check
    int sum = 0;
    for(int i = 0; i < N; i++) {     
        //printf("%i %i\n", h_out[i], h_checkResults[i]);
        sum += h_checkResults[i] - h_out[i];
    }

    cout << "Error between CPU and GPU: " << sum << endl;

    delete[] h_in;
    delete[] h_out;
    delete[] h_checkResults;

    return 0;
}

运行此类示例所需的Utilities.cu和Utilities.cuh文件在此github页面上维护.

Visual Profiler时间轴显示CUDA Thrust操作和内存传输的并发性

Answer 2

不,你没有遗漏任何东西(至少在CUDA 6.0附带的发布快照中).

最初的基于Thrust标签的调度系统故意将所有潜在的CUDA API调用抽象出去,牺牲了一些易用性和一致性的性能(请记住,推力除了CUDA之外还有后端).如果您需要这种灵活性,则需要尝试其他库(例如CUB).

在自CUDA 7.0快照以来的版本中,可以通过执行策略和调度功能为推力操作设置选择流.