我正在收到错误:
文件中的Cufft错误
我使用这个文件是为了加载FFT并将它们传递给另一个文件。
//----function to check for errors-------------------------------------------------
#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,"\nGPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
//function to check for cuFFT errors --------------------------------------------------
#define CUFFT_SAFE_CALL( call) do { \
cufftResult err = call; \
if (err != CUFFT_SUCCESS) { \
fprintf(stderr, "Cufft error in file '%s' in line %i : %s.\n", \
__FILE__, __LINE__, "error" ); \
exit(EXIT_FAILURE); \
} \
} while (0)
#define NX 128*128
#define NY 16
#define BATCH 16
#define NRANK 2
void FFT_transform(cufftDoubleComplex** B_in)
{
int n[NRANK] = {NX, NY};
//size of B
int Bsize=NX*NY*BATCH;
//allocate host memory
*B_in=(cufftDoubleComplex*)malloc(Bsize*sizeof(cufftDoubleComplex));
for (int i=0;i<NX*NY;i++){
for (int j=0;j<BATCH;j++){
(*B_in)[i*BATCH+j].x=(i*BATCH+j)*2;
(*B_in)[i*BATCH+j].y=(i*BATCH+j)*2+1;
}
}
//allocate device memory
cufftDoubleComplex* B_dev;
gpuErrchk(cudaMalloc((void**) &B_dev,Bsize* sizeof(cufftDoubleComplex)));
if (cudaGetLastError() != cudaSuccess){
fprintf(stderr, "Cuda error: Failed to allocate\n");
return;
}
// copy arrays from host to device
gpuErrchk(cudaMemcpy(B_dev, *B_in,Bsize* sizeof(cufftDoubleComplex), cudaMemcpyHostToDevice));
// Create a 2D FFT plan
cufftHandle plan;
CUFFT_SAFE_CALL(cufftPlan2d(&plan,NX,NY,CUFFT_Z2Z));
if (cufftPlanMany(&plan, NRANK, n,NULL, 1, 0,NULL, 1, 0,CUFFT_Z2Z,BATCH) != CUFFT_SUCCESS){
fprintf(stderr, "CUFFT Error: Unable to create plan\n");
return;
}
if (cufftSetCompatibilityMode(plan, CUFFT_COMPATIBILITY_NATIVE)!= CUFFT_SUCCESS){
fprintf(stderr, "CUFFT Error: Unable to set compatibility mode to native\n");
return;
}
// perform transform
CUFFT_SAFE_CALL(cufftExecZ2Z(plan,(cufftDoubleComplex *)(*B_in), (cufftDoubleComplex *)B_dev, CUFFT_FORWARD));
if (cufftExecZ2Z(plan,*B_in,B_dev,CUFFT_FORWARD) != CUFFT_SUCCESS){
fprintf(stderr, "CUFFT Error: Unable to execute plan\n");
return;
}
if (cudaThreadSynchronize() != cudaSuccess){
fprintf(stderr, "Cuda error: Failed to synchronize\n");
return;
}
// copy result from device to host
gpuErrchk(cudaMemcpy(*B_in, B_dev,Bsize*sizeof(cufftDoubleComplex), cudaMemcpyDeviceToHost));
//Destroy CUFFT context
CUFFT_SAFE_CALL(cufftDestroy(plan));
//clean up device memory
gpuErrchk(cudaFree(B_dev));
}我正在收到以下错误:
CUFFT_SAFE_CALL(cufftExecZ2Z(plan,(cufftDoubleComplex *)(*B_in), (cufftDoubleComplex *)B_dev, CUFFT_FORWARD));发布于 2014-02-24 10:48:08
您将得到错误,因为B_in是指向主机内存的指针,而不是指向设备内存的指针,这是非法的。在CUFFT中,输入总是在设备内存中。在执行转换之前,您需要使用cudaMemcpy将B_in的内容传输到B_dev,然后提供B_dev作为输入和输出,这将导致就地转换。CUFFT文档这里清楚地描述了这一点。
https://stackoverflow.com/questions/21984447
复制相似问题