cuFFT和流

Question

cuFFT和流

我正在尝试使用流异步启动多个CUDA FFT内核.为此,我正在创建我的流,cuFFT前向和反向计划如下:

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)
}

Run Code Online (Sandbox Code Playgroud)

在main函数中,我正在启动正向FFT,如下所示:

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

Run Code Online (Sandbox Code Playgroud)

我还有其他内核,我使用相同的流异步启动.

当我使用Visual Profiler 5.0分析我的应用程序时,我发现除了CUDA FFT(正向和反向)之外的所有内核并行运行并重叠.FFT内核确实在不同的流中运行,但它们不重叠,因为它们实际上是顺序运行的.谁能告诉我我的问题是什么？

我的环境是VS 2008,64位,Windows 7.

谢谢.

Answer 1

Jac*_*ern 6

这是在Kepler体系结构中使用CUDA中的流的cuFFT执行和memcopies的工作示例.

这是代码:

#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;
}

Run Code Online (Sandbox Code Playgroud)

请根据CUFFT错误处理添加cuFFT错误检查.

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

N = 5000

在此输入图像描述