How to perform a Real to Complex Transformation with cuFFT

问题

The following code has been adapted from here to apply to a single 1D transformation using cufftPlan1d. Ultimately I want to perform a batched in place R2C transformation, but code below perfroms a single transformation using a separate input and output array.

How can adapt this code to perform a the transformation inplace, therefore reducing the amount of memory allocated on the device?

Thanks
Cuda 6.5 - Note: I'm running the code from a mexFunction in MATLAB 2015a

Code:

#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));

}

回答1:

The solution has already been given in another answer: https://stackoverflow.com/a/19208070/678093

For your example, this means:

Allocate input as cufftComplex:

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

In-Place transformation:

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

btw: MATLAB also contains a GPU accelerated version of fft(), maybe this could be useful for you as well: http://de.mathworks.com/help/distcomp/run-built-in-functions-on-a-gpu.html#btjw5gk

回答2:

Here is my own complete solution that start with cufftReal instead

void process(double *x, double *y, size_t n){
// --- Host side input data allocation and initialization
cufftReal *hostInputData = (cufftReal*)malloc(DATASIZE*sizeof(cufftReal));
for (int j=0; j<DATASIZE; j++) hostInputData[j] = (cufftReal)x[j];

// --- Device side input data allocation and initialization
cufftReal *deviceData; 
gpuErrchk(cudaMalloc((void**)&deviceData, (DATASIZE / 2 + 1) * BATCH * sizeof(cufftComplex)));
cudaMemcpy(deviceData, hostInputData, DATASIZE * sizeof(cufftReal), cudaMemcpyHostToDevice);

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

cufftResult cufftStatus;
cufftHandle handle;

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

cufftStatus = cufftExecR2C(handle,  deviceData, (cufftComplex*)deviceData);
if (cufftStatus != cudaSuccess) { mexPrintf("cufftExecR2C failed!"); }  

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

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

cufftDestroy(handle);
gpuErrchk(cudaFree(deviceData));}

来源：https://stackoverflow.com/questions/29240591/how-to-perform-a-real-to-complex-transformation-with-cufft

标签

cuda

cufft