c - 如何使用 cuFFT 执行实数到复数的转换

Question

以下代码改编自here使用 cufftPlan1d 应用于单个一维变换。最终我想执行批处理就地 R2C 转换，但下面的代码使用单独的输入和输出数组执行单个转换。

如何调整此代码以就地执行转换，从而减少设备上分配的内存量？

谢谢
Cuda 6.5 - 注意:我正在从 MATLAB 2015a 中的 mexFunction 运行代码

代码:

#include <stdlib.h>
#include <stdio.h>
#include <cuda_runtime.h>
#include <cufft.h>
#define DATASIZE 8
#define BATCH 1
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const 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);
   }
}

void main(int argc, char **argv)
{   

// --- Host side input data allocation and initialization
cufftReal *hostInputData = (cufftReal*)malloc(DATASIZE*sizeof(cufftReal));
for (int j=0; j<DATASIZE; j++) hostInputData[j] = (cufftReal)(j + 1);

// --- Device side input data allocation and initialization
cufftReal *deviceInputData; 
gpuErrchk(cudaMalloc((void**)&deviceInputData, DATASIZE * sizeof(cufftReal)));
cudaMemcpy(deviceInputData, hostInputData, DATASIZE * sizeof(cufftReal), cudaMemcpyHostToDevice);

// --- Host side output data allocation
cufftComplex *hostOutputData = (cufftComplex*)malloc((DATASIZE / 2 + 1) * BATCH * sizeof(cufftComplex));

// --- Device side output data allocation
cufftComplex *deviceOutputData;   gpuErrchk(cudaMalloc((void**)&deviceOutputData, (DATASIZE / 2 + 1) * sizeof(cufftComplex)));

cufftResult cufftStatus;
cufftHandle handle;

cufftStatus = cufftPlan1d(&handle, DATASIZE, CUFFT_R2C, BATCH);
if (cufftStatus != cudaSuccess) { mexPrintf("cufftPlan1d failed!"); }       

cufftStatus = cufftExecR2C(handle,  deviceInputData, deviceOutputData);
if (cufftStatus != cudaSuccess) { mexPrintf("cufftExecR2C failed!"); }  

// --- Device->Host copy of the results
gpuErrchk(cudaMemcpy(hostOutputData, deviceOutputData, (DATASIZE / 2 + 1) * sizeof(cufftComplex), cudaMemcpyDeviceToHost));

for (int j=0; j<(DATASIZE / 2 + 1); j++)
        printf("%i %f %f\n", j, hostOutputData[j].x, hostOutputData[j].y);

cufftDestroy(handle);
gpuErrchk(cudaFree(deviceOutputData));
gpuErrchk(cudaFree(deviceInputData));

}

Answer 1

解决方案已经在另一个答案中给出:https://stackoverflow.com/a/19208070/678093

对于您的示例，这意味着:

将输入分配为 cufftComplex:

cufftComplex *deviceInputData;
gpuErrchk(cudaMalloc((void**)&deviceInputData, DATASIZE * sizeof(cufftComplex)));
cudaMemcpy(deviceInputData, hostInputData, DATASIZE * sizeof(cufftReal), cudaMemcpyHostToDevice);

就地转换:

cufftStatus = cufftExecR2C(handle,  (cufftReal *)deviceInputData, deviceInputData);
gpuErrchk(cudaMemcpy(hostOutputData, deviceInputData, (DATASIZE / 2 + 1) * sizeof(cufftComplex), cudaMemcpyDeviceToHost));

顺便说一句:MATLAB 还包含 fft() 的 GPU 加速版本，也许这对您也有用:http://de.mathworks.com/help/distcomp/run-built-in-functions-on-a-gpu.html#btjw5gk