cuda

Question

我在使用cufftPlanMany时遇到了麻烦。在创建计划并进行正向和反向FFT后，我无法获得原始数据。请查找附件中的代码的最低版本。

program test_cufft
  use cudafor
  use cufft

  integer :: plan_r2c
  integer :: plan_c2r
  real,allocatable,dimension(:,:,:,:), device :: eta_d
  complex,allocatable,dimension(:,:,:,:), device :: etak_d

  nv = 4
  nx = 256
  ny = 512
  nz = 512
  nx21 = nx/2+1

  allocate( eta_d(nv,nx,ny,nz) )
  allocate( etak_d(nv,nx21,ny,nz) )

  batch = nv;
  rank = 3;
  n = (/ nx, ny, nz /);
  idist = nx*ny*nz;
  odist = nx21*ny*nz;
  inembed = (/ nx, ny, nz /);
  onembed = (/ nx21, ny, nz /);
  istride = 1;
  ostride = 1;

  istat = cufftPlanMany( plan_r2c, rank, n, inembed, istride, idist, &
                         onembed, ostride, odist, CUFFT_R2C, batch )
  istat = cufftPlanMany( plan_c2r, rank, n, onembed, ostride, odist, &
                         inembed, istride, idist, CUFFT_C2R, batch )

  ! Initialize eta_d

  istat = cufftExecR2C( plan_r2c, eta_d, etak_d )
  istat = cufftExecC2R( plan_c2r, etak_d, eta_d )
  eta_d = eta_d/idist

end program test_cufft

问题是，在我做了正向和反向的FFTs之后，我无法得到原始数据。拜托，我做错什么了？数据的排序应该是eta_d(batch,nx,ny,nz) or eta_d(nx,ny,nz,batch)吗

Answer 1

我认为正确的排序是(nz, ny, nx, batch)

但是，将它们与数组索引和存储顺序联系起来也很重要。

在CUFFT 术语中，对于三维变换(*)，nz方向是变化最快的索引，其中典型用法(stride=1)是内存中的相邻数据，对应于变换中的相邻元素。

对于R2C/C2R转换类型，这个方向(我认为它是一行上的元素，即"z“索引是列索引)也是在复杂域中得到”约简“的多维转换的方向。

考虑到这一点，我将以如下方式重写您的代码：

$ cat t4.cuf
program test_cufft
  use cudafor
  use cufft

  integer :: plan_r2c
  integer :: plan_c2r
  real,allocatable,dimension(:,:,:,:), managed :: eta_d
  complex,allocatable,dimension(:,:,:,:), managed :: etak_d
  integer :: n(3), inembed(3), onembed(3),rank,istride,idist,ostride,odist,batch

  nv = 4
  nx = 8
  ny = 8
  nz = 4
  nz21 = nz/2+1

  allocate( eta_d(nz,ny,nx,nv) )
  allocate( etak_d(nz21,ny,nx,nv) )

  batch = nv;
  rank = 3;
  n = (/ nx, ny, nz /);
  idist = nx*ny*nz;
  odist = nx*ny*nz21;
  inembed = (/ nx, ny, nz /);
  onembed = (/ nx, ny, nz21 /);
  istride = 1;
  ostride = 1;

  istat = cufftPlanMany( plan_r2c, rank, n, inembed, istride, idist, onembed, ostride, odist, CUFFT_R2C, batch )
  istat = cufftPlanMany( plan_c2r, rank, n, onembed, ostride, odist, inembed, istride, idist, CUFFT_C2R, batch )

  ! Initialize eta_d
  eta_d(:,:,:,:) = 1.0
  eta_d(1,1,1,2) = 2.0
  istat = cufftExecR2C( plan_r2c, eta_d, etak_d )
  istat = cudaDeviceSynchronize()
  eta_d(:,:,:,:) = 0.0
  istat = cufftExecC2R( plan_c2r, etak_d, eta_d )
  istat = cudaDeviceSynchronize()
  eta_d = eta_d/idist
  print *,eta_d(1,1,1,1)
  print *,eta_d(1,1,1,2)
end program test_cufft
$ nvfortran t4.cuf -lcufft
$ ./a.out
    1.000000
    2.000000
$

(NVIDIA HPC SDK 20.9，Tesla V100 GPU)

它似乎给出了我的简单测试用例的预期结果。

(*)对于二维变换，ny维数变化最快，对于一维变换，nx维数(当然)变化最快。

多维转换和先进的数据布局CUFFT手册的章节也可能是有用的阅读。