cuda fortran cufftPlanMany
cuda fortran cufftPlanMany
我在使用 cufftPlanMany 时遇到问题。创建计划并进行正向和反向 FFT 后，我无法取回原始数据。请查找随附的代码的最低版本。
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FFT之后，我无法取回原始数据。
请问，我做错了什么？
数据的顺序应该是eta_d(batch,nx,ny,nz) or eta_d(nx,ny,nz,batch)?
我会说正确的顺序是 (nz, ny, nx, batch)
但将这些与您的数组索引和存储顺序相关联也很重要。
在 CUFFT terminology 中，对于 3D 变换 (*)，nz 方向是变化最快的索引，典型用法 (stride=1) 是内存中的相邻数据，对应于相邻转换中的元素。
这个方向（我认为它是沿着一行的元素，即“z”索引是列索引）也是在复杂域中“减少”的多维变换的方向，对于R2C/C2R 转换类型。
考虑到这一点，我会这样重写您的代码：
$ 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）
它似乎为我的简单测试用例提供了预期的结果。
(*) 对于二维变换，ny 维度变化最快，而对于一维变换，nx 维度（当然）变化最快。
多维变换和高级数据布局sections of the CUFFT manual也可能是有用的读物。
我在使用 cufftPlanMany 时遇到问题。创建计划并进行正向和反向 FFT 后，我无法取回原始数据。请查找随附的代码的最低版本。
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FFT之后，我无法取回原始数据。
请问，我做错了什么？
数据的顺序应该是eta_d(batch,nx,ny,nz) or eta_d(nx,ny,nz,batch)?
我会说正确的顺序是 (nz, ny, nx, batch)
但将这些与您的数组索引和存储顺序相关联也很重要。
在 CUFFT terminology 中，对于 3D 变换 (*)，nz 方向是变化最快的索引，典型用法 (stride=1) 是内存中的相邻数据，对应于相邻转换中的元素。
这个方向（我认为它是沿着一行的元素，即“z”索引是列索引）也是在复杂域中“减少”的多维变换的方向，对于R2C/C2R 转换类型。
考虑到这一点，我会这样重写您的代码：
$ 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）
它似乎为我的简单测试用例提供了预期的结果。
(*) 对于二维变换，ny 维度变化最快，而对于一维变换，nx 维度（当然）变化最快。
多维变换和高级数据布局sections of the CUFFT manual也可能是有用的读物。