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根據該文檔時，cufftSetStream()功能cufftSetStream導致垃圾輸出。難道我做錯了什麼？
相關聯的CUDA流與CUFFT計劃。在計劃執行期間所做的所有內核啓動現在都通過關聯的流[...直到...]完成，流將通過另一次調用cufftSetStream（）進行更改。
不幸的是，結果變成垃圾。下面是一個例子，它通過兩種方式執行一系列轉換來演示：一種是每個流有自己的專用計劃，另一種是單個計劃被重用，如上面的文檔所示。前者的行爲如預期，重用/ cufftSetStream方法在大多數轉換中都存在錯誤。我在CentOS 7 linux上試用過的兩張顯卡（GTX 750 ti，Titan X）上觀察到了這種情況，其中有 Cuda編譯工具，7.0版，V7.0.27;併發布7.5，V7.5.17。
編輯：請參閱下面的「FIX」註釋以解決某些問題。
#include <cufft.h>
#include <stdexcept>
#include <iostream>
#include <numeric>
#include <vector>
#define ck(cmd) if (cmd) { std::cerr << "error at line " << __LINE__ << std::endl;exit(1);}
__global__
void fill_input(cufftComplex * buf, int batch,int nbins,int stride,int seed)
{
for (int i = blockDim.y * blockIdx.y + threadIdx.y; i< batch;i += gridDim.y*blockDim.y)
for (int j = blockDim.x * blockIdx.x + threadIdx.x; j< nbins;j += gridDim.x*blockDim.x)
buf[i*stride + j] = make_cuFloatComplex((i+seed)%101 - 50,(j+seed)%41-20);
}
__global__
void check_output(const float * buf1,const float * buf2,int batch, int nfft, int stride, int * errors)
{
for (int i = blockDim.y * blockIdx.y + threadIdx.y; i< batch;i += gridDim.y*blockDim.y) {
for (int j = blockDim.x * blockIdx.x + threadIdx.x; j< nfft;j += gridDim.x*blockDim.x) {
float e=buf1[i*stride+j] - buf2[i*stride+j];
if (e*e > 1) // gross error
atomicAdd(errors,1);
}
}
}
void demo(bool reuse_plan)
{
if (reuse_plan)
std::cout << "Reusing the same fft plan with multiple stream via cufftSetStream ... ";
else
std::cout << "Giving each stream its own dedicated fft plan ... ";
int nfft = 1024;
int batch = 1024;
int nstreams = 8;
int nbins = nfft/2+1;
int nit=100;
size_t inpitch,outpitch;
std::vector<cufftComplex*> inbufs(nstreams);
std::vector<float*> outbufs(nstreams);
std::vector<float*> checkbufs(nstreams);
std::vector<cudaStream_t> streams(nstreams);
std::vector<cufftHandle> plans(nstreams);
for (int i=0;i<nstreams;++i) {
ck(cudaStreamCreate(&streams[i]));
ck(cudaMallocPitch((void**)&inbufs[i],&inpitch,nbins*sizeof(cufftComplex),batch));
ck(cudaMallocPitch((void**)&outbufs[i],&outpitch,nfft*sizeof(float),batch));
ck(cudaMallocPitch((void**)&checkbufs[i],&outpitch,nfft*sizeof(float),batch));
if (i==0 || reuse_plan==false)
ck (cufftPlanMany(&plans[i],1,&nfft,&nbins,1,inpitch/sizeof(cufftComplex),&nfft,1,outpitch/sizeof(float),CUFFT_C2R,batch));
}
// fill the input buffers and FFT them to get a baseline for comparison
for (int i=0;i<nstreams;++i) {
fill_input<<<20,dim3(32,32)>>>(inbufs[i],batch,nbins,inpitch/sizeof(cufftComplex),i);
ck (cudaGetLastError());
if (reuse_plan) {
ck (cufftExecC2R(plans[0],inbufs[i],checkbufs[i]));
}else{
ck (cufftExecC2R(plans[i],inbufs[i],checkbufs[i]));
ck(cufftSetStream(plans[i],streams[i])); // only need to set the stream once
}
ck(cudaDeviceSynchronize());
}
// allocate a buffer for the error count
int * errors;
cudaMallocHost((void**)&errors,sizeof(int)*nit);
memset(errors,0,sizeof(int)*nit);
/* FIX: an event can protect the plan internal buffers
by serializing access to the plan
cudaEvent_t ev;
cudaEventCreateWithFlags(&ev,cudaEventDisableTiming);
*/
// perform the FFTs and check the outputs on streams
for (int it=0;it<nit;++it) {
int k = it % nstreams;
ck(cudaStreamSynchronize(streams[k])); // make sure any prior kernels have completed
if (reuse_plan) {
// FIX: ck(cudaStreamWaitEvent(streams[k],ev,0));
ck(cufftSetStream(plans[0],streams[k]));
ck(cufftExecC2R(plans[0],inbufs[k],outbufs[k]));
// FIX: ck(cudaEventRecord(ev,streams[k]));
}else{
ck(cufftExecC2R(plans[k],inbufs[k],outbufs[k]));
}
check_output<<<100,dim3(32,32),0,streams[k]>>>(outbufs[k],checkbufs[k],batch,nfft,outpitch/sizeof(float),&errors[it]);
ck (cudaGetLastError());
}
ck(cudaDeviceSynchronize());
// report number of errors
int errcount=0;
for (int it=0;it<nit;++it)
if (errors[it])
++errcount;
std::cout << errcount << " of " << nit << " transforms had errors\n";
for (int i=0;i<nstreams;++i) {
cudaFree(inbufs[i]);
cudaFree(outbufs[i]);
cudaStreamDestroy(streams[i]);
if (i==0 || reuse_plan==false)
cufftDestroy(plans[i]);
}
}
int main(int argc,char ** argv)
{
demo(false);
demo(true);
return 0;
}
典型輸出
給予每個流其自己的專用FFT計劃... 0 100的變換有錯誤
重用經由cufftSetStream多流相同的FFT計劃... 87 100轉換有錯誤
當我編譯並運行你張貼在一個很不起眼的移動GPU使用CUDA 7.0的代碼，我得到了這兩種情況下0錯誤。 – talonmies
@talonmies，感謝您的數據點。我只用cuda 7.0試過 - 也失敗了（見編輯）也許卡的「謙虛」阻止它失敗（即更少的資源==更少的競爭條件）。你在什麼操作系統上？ –
帶有計算2.1設備的Windows 10 – talonmies