首页
学习
活动
专区
圈层
工具
发布

cuFFT和流
EN

Stack Overflow用户
提问于 2013-06-08 01:44:04
回答 3查看 4.6K关注 0票数 8

我正在尝试使用streams异步启动多个CUDA FFT内核。为此,我创建了我的streams、cuFFT正向和反向计划,如下所示:

代码语言:javascript
复制
streams = (cudaStream_t*) malloc(sizeof(cudaStream_t)*streamNum);
plansF = (cufftHandle *) malloc(sizeof(cufftHandle)*streamNum);
plansI = (cufftHandle *) malloc(sizeof(cufftHandle)*streamNum);
for(int i=0; i<streamNum; i++)  
{
    cudaStreamCreate(&streams[i]);
    CHECK_ERROR(5)
    cufftPlan1d(&plansF[i], ticks, CUFFT_R2C,1);
    CHECK_ERROR(5)
    cufftPlan1d(&plansI[i], ticks, CUFFT_C2R,1);
    CHECK_ERROR(5)
    cufftSetStream(plansF[i],streams[i]);
    CHECK_ERROR(5)
    cufftSetStream(plansI[i],streams[i]);
    CHECK_ERROR(5)
}

main函数中,我按如下方式启动转发FFT:

代码语言:javascript
复制
for(w=1;w<q;w++)
  {
    cufftExecR2C(plansF[w], gpuMem1+k,gpuMem2+j);
    CHECK_ERROR(8)
    k += rect_small_real;
    j += rect_small_complex;
  }

我还使用相同的流异步启动了其他内核。

当我使用Visual Profiler 5.0分析我的应用程序时,我看到除了CUDA FFT (正向和反向)之外的所有内核都是并行和重叠运行的。FFT内核确实在不同的流中运行,但它们不会重叠,因为它们实际上是按顺序运行的。有人能告诉我我的问题出在哪里吗?

我的环境是VS2008,64位,Windows 7。

谢谢。

EN

回答 3

Stack Overflow用户

发布于 2014-08-05 04:58:06

这是一个在cuFFT架构上使用CUDA中的流执行和memcopies的工作示例。

代码如下:

代码语言:javascript
复制
#include <stdio.h>

#include <cufft.h>

#define NUM_STREAMS 3

/********************/
/* CUDA ERROR CHECK */
/********************/
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
   if (code != cudaSuccess) 
   {
      fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
      if (abort) exit(code);
   }
}

/********/
/* MAIN */
/********/
int main()
{
    const int N = 5000;

    // --- Host input data initialization
    float2 *h_in1 = new float2[N];
    float2 *h_in2 = new float2[N];
    float2 *h_in3 = new float2[N];
    for (int i = 0; i < N; i++) {
        h_in1[i].x = 1.f;
        h_in1[i].y = 0.f;
        h_in2[i].x = 1.f;
        h_in2[i].y = 0.f;
        h_in3[i].x = 1.f;
        h_in3[i].y = 0.f;
    }

    // --- Host output data initialization
    float2 *h_out1 = new float2[N];
    float2 *h_out2 = new float2[N];
    float2 *h_out3 = new float2[N];
    for (int i = 0; i < N; i++) {
        h_out1[i].x = 0.f;
        h_out1[i].y = 0.f;
        h_out2[i].x = 0.f;
        h_out2[i].y = 0.f;
        h_out3[i].x = 0.f;
        h_out3[i].y = 0.f;
    }

    // --- Registers host memory as page-locked (required for asynch cudaMemcpyAsync)
    gpuErrchk(cudaHostRegister(h_in1, N*sizeof(float2), cudaHostRegisterPortable));
    gpuErrchk(cudaHostRegister(h_in2, N*sizeof(float2), cudaHostRegisterPortable));
    gpuErrchk(cudaHostRegister(h_in3, N*sizeof(float2), cudaHostRegisterPortable));
    gpuErrchk(cudaHostRegister(h_out1, N*sizeof(float2), cudaHostRegisterPortable));
    gpuErrchk(cudaHostRegister(h_out2, N*sizeof(float2), cudaHostRegisterPortable));
    gpuErrchk(cudaHostRegister(h_out3, N*sizeof(float2), cudaHostRegisterPortable));

    // --- Device input data allocation
    float2 *d_in1;          gpuErrchk(cudaMalloc((void**)&d_in1, N*sizeof(float2)));
    float2 *d_in2;          gpuErrchk(cudaMalloc((void**)&d_in2, N*sizeof(float2)));
    float2 *d_in3;          gpuErrchk(cudaMalloc((void**)&d_in3, N*sizeof(float2)));
    float2 *d_out1;         gpuErrchk(cudaMalloc((void**)&d_out1, N*sizeof(float2)));
    float2 *d_out2;         gpuErrchk(cudaMalloc((void**)&d_out2, N*sizeof(float2)));
    float2 *d_out3;         gpuErrchk(cudaMalloc((void**)&d_out3, N*sizeof(float2)));

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

    // --- Creates cuFFT plans and sets them in streams
    cufftHandle* plans = (cufftHandle*) malloc(sizeof(cufftHandle)*NUM_STREAMS);
    for (int i = 0; i < NUM_STREAMS; i++) {
        cufftPlan1d(&plans[i], N, CUFFT_C2C, 1);
        cufftSetStream(plans[i], streams[i]);
    }

    // --- Async memcopyes and computations
    gpuErrchk(cudaMemcpyAsync(d_in1, h_in1, N*sizeof(float2), cudaMemcpyHostToDevice, streams[0]));
    gpuErrchk(cudaMemcpyAsync(d_in2, h_in2, N*sizeof(float2), cudaMemcpyHostToDevice, streams[1]));
    gpuErrchk(cudaMemcpyAsync(d_in3, h_in3, N*sizeof(float2), cudaMemcpyHostToDevice, streams[2]));
    cufftExecC2C(plans[0], (cufftComplex*)d_in1, (cufftComplex*)d_out1, CUFFT_FORWARD);
    cufftExecC2C(plans[1], (cufftComplex*)d_in2, (cufftComplex*)d_out2, CUFFT_FORWARD);
    cufftExecC2C(plans[2], (cufftComplex*)d_in3, (cufftComplex*)d_out3, CUFFT_FORWARD);
    gpuErrchk(cudaMemcpyAsync(h_out1, d_out1, N*sizeof(float2), cudaMemcpyDeviceToHost, streams[0]));
    gpuErrchk(cudaMemcpyAsync(h_out2, d_out2, N*sizeof(float2), cudaMemcpyDeviceToHost, streams[1]));
    gpuErrchk(cudaMemcpyAsync(h_out3, d_out3, N*sizeof(float2), cudaMemcpyDeviceToHost, streams[2]));

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

    // --- Releases resources
    gpuErrchk(cudaHostUnregister(h_in1));
    gpuErrchk(cudaHostUnregister(h_in2));
    gpuErrchk(cudaHostUnregister(h_in3));
    gpuErrchk(cudaHostUnregister(h_out1));
    gpuErrchk(cudaHostUnregister(h_out2));
    gpuErrchk(cudaHostUnregister(h_out3));
    gpuErrchk(cudaFree(d_in1));
    gpuErrchk(cudaFree(d_in2));
    gpuErrchk(cudaFree(d_in3));
    gpuErrchk(cudaFree(d_out1));
    gpuErrchk(cudaFree(d_out2));
    gpuErrchk(cudaFree(d_out3));

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

    delete[] h_in1;
    delete[] h_in2;
    delete[] h_in3;
    delete[] h_out1;
    delete[] h_out2;
    delete[] h_out3;

    cudaDeviceReset();  

    return 0;
}

请根据CUFFT error handling添加cuFFT错误检查。

下面,提供了在开普勒K20c卡上测试上述算法时的一些分析信息。正如您将看到的,只有在拥有足够大的N的情况下,您才能在计算和内存传输之间实现真正的重叠。

N = 5000

N = 50000

N = 500000

票数 7
EN

Stack Overflow用户

发布于 2013-11-27 21:20:46

问题出在您使用的硬件上。

所有支持CUDA的GPU都能够同时以两种方式执行内核和复制数据。然而,只有具有计算能力3.5的设备才具有名为Hyper-Q的功能。

简而言之,在这些GPU中实现了几个(我认为是16个)硬件内核队列。在以前的GPU中,一个硬件队列是可用的。

这意味着cudaStreams只是虚拟的,它们对旧硬件的使用只有在重叠计算和内存复制的情况下才有意义。当然,这不仅适用于cuFFT,也适用于您自己的内核!

请深入查看visual profiler的输出。您可能会无意中将时间线可视化视为GPU执行的确切数据。然而,事情并没有那么简单。有几行显示的数据可能指的是内核启动行执行的时间点(通常是橙色的)。这一行对应于特定内核在GPU上的执行(蓝色矩形)。内存传输也是如此(确切的时间显示为浅棕色的矩形)。

霍普我帮你解决了你的问题

票数 2
EN

Stack Overflow用户

发布于 2016-11-02 01:35:52

这里有一个@JackOLantern代码的摘要,它允许简单地改变FFT的数量,FFT长度和流计数,以在nvvp中试验GPU利用率。

代码语言:javascript
复制
// Compile with:
// nvcc --std=c++11 stream_parallel.cu -o stream_parallel -lcufft

#include <iostream>

#include <cuda.h>
#include <cuda_runtime.h>

#include <cufft.h>

// Print file name, line number, and error code when a CUDA error occurs.
#define check_cuda_errors(val)  __check_cuda_errors__ ( (val), #val, __FILE__, __LINE__ )

template <typename T>
inline void __check_cuda_errors__(T code, const char *func, const char *file, int line) {
    if (code) {
    std::cout << "CUDA error at "
          << file << ":" << line << std::endl
          << "error code: " << (unsigned int) code
          << " type: \""  << cudaGetErrorString(cudaGetLastError()) << "\"" << std::endl
          << "func: \"" << func << "\""
          << std::endl;
    cudaDeviceReset();
    exit(EXIT_FAILURE);
    }
}

int main(int argc, char *argv[]) {

    // Number of FFTs to compute.
    const int NUM_DATA = 64;

    // Length of each FFT.
    const int N = 1048576;

    // Number of GPU streams across which to distribute the FFTs.
    const int NUM_STREAMS = 4;

    // Allocate and initialize host input data.
    float2 **h_in = new float2 *[NUM_STREAMS];
    for (int ii = 0; ii < NUM_STREAMS; ii++) {
        h_in[ii] = new float2[N];
        for (int jj = 0; jj < N; ++jj) {
            h_in[ii][jj].x = (float) 1.f;
            h_in[ii][jj].y = (float) 0.f;
        }
    }

    // Allocate and initialize host output data.
    float2 **h_out = new float2 *[NUM_STREAMS];
    for (int ii = 0; ii < NUM_STREAMS; ii++) {
    h_out[ii] = new float2[N];
    for (int jj = 0; jj < N; ++jj) {
            h_out[ii][jj].x = 0.f;
            h_out[ii][jj].y = 0.f;
        }
    }

    // Pin host input and output memory for cudaMemcpyAsync.
    for (int ii = 0; ii < NUM_STREAMS; ii++) {
        check_cuda_errors(cudaHostRegister(h_in[ii], N*sizeof(float2), cudaHostRegisterPortable));
        check_cuda_errors(cudaHostRegister(h_out[ii], N*sizeof(float2), cudaHostRegisterPortable));
    }

    // Allocate pointers to device input and output arrays.
    float2 **d_in = new float2 *[NUM_STREAMS];
    float2 **d_out = new float2 *[NUM_STREAMS];

    // Allocate intput and output arrays on device.
    for (int ii = 0; ii < NUM_STREAMS; ii++) {
        check_cuda_errors(cudaMalloc((void**)&d_in[ii], N*sizeof(float2)));
        check_cuda_errors(cudaMalloc((void**)&d_out[ii], N*sizeof(float2)));
    }

    // Create CUDA streams.
    cudaStream_t streams[NUM_STREAMS];
    for (int ii = 0; ii < NUM_STREAMS; ii++) {
        check_cuda_errors(cudaStreamCreate(&streams[ii]));
    }

    // Creates cuFFT plans and sets them in streams
    cufftHandle* plans = (cufftHandle*) malloc(sizeof(cufftHandle)*NUM_STREAMS);
    for (int ii = 0; ii < NUM_STREAMS; ii++) {
        cufftPlan1d(&plans[ii], N, CUFFT_C2C, 1);
        cufftSetStream(plans[ii], streams[ii]);
    }

    // Fill streams with async memcopies and FFTs.
    for (int ii = 0; ii < NUM_DATA; ii++) {
        int jj = ii % NUM_STREAMS;
        check_cuda_errors(cudaMemcpyAsync(d_in[jj], h_in[jj], N*sizeof(float2), cudaMemcpyHostToDevice, streams[jj]));
        cufftExecC2C(plans[jj], (cufftComplex*)d_in[jj], (cufftComplex*)d_out[jj], CUFFT_FORWARD);
        check_cuda_errors(cudaMemcpyAsync(h_out[jj], d_out[jj], N*sizeof(float2), cudaMemcpyDeviceToHost, streams[jj]));
    }

    // Wait for calculations to complete.
    for(int ii = 0; ii < NUM_STREAMS; ii++) {
        check_cuda_errors(cudaStreamSynchronize(streams[ii]));
    }

    // Free memory and streams.
    for (int ii = 0; ii < NUM_STREAMS; ii++) {
        check_cuda_errors(cudaHostUnregister(h_in[ii]));
        check_cuda_errors(cudaHostUnregister(h_out[ii]));
        check_cuda_errors(cudaFree(d_in[ii]));
        check_cuda_errors(cudaFree(d_out[ii]));
        delete[] h_in[ii];
        delete[] h_out[ii];
        check_cuda_errors(cudaStreamDestroy(streams[ii]));
    }

    delete plans;

    cudaDeviceReset();  

    return 0;
}
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/16990025

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档