Input Fortran:读取输入
我正在尝试运行一个非常大的FORTRAN模型,所以我不能给出所有涉及的代码,但我希望我能提供足够的信息,使其有意义。 代码编译良好(使用英特尔2016.0.109编译器、OpenMPI 1.10.2和HDF5 1.8.17) 当我尝试运行它时,它告诉我我的两个输入(称为NZG和NZS)被设置为-999,因此它失败了Input Fortran:读取输入,input,fortran,intel-fortran,Input,Fortran,Intel Fortran,我正在尝试运行一个非常大的FORTRAN模型,所以我不能给出所有涉及的代码,但我希望我能提供足够的信息,使其有意义。 代码编译良好(使用英特尔2016.0.109编译器、OpenMPI 1.10.2和HDF5 1.8.17) 当我尝试运行它时,它告诉我我的两个输入(称为NZG和NZS)被设置为-999,因此它失败了 > >>>> opspec_grid error! in your namelist! > ---> Reason:
> >>>> opspec_grid error! in your namelist!
> ---> Reason: Too few soil layers. Set it to at least 2. Your nzg is currently set to -999...
> ---> Reason: Too few maximum # of snow layers. Set it to at least 1. Your nzs is currently set to -999.
但是,在输入文件中,它们实际上被指定为
NL%NZG = 9
NL%NZS = 1
我浏览了所有与这些变量有关的模块,但找不到任何应该偏离轨道的地方。
所以我现在开始想,可能是读取值的方式有问题。输入文件是一个文本文件。变量在读取它们的模块中指定为整数,仅供参考。
我应该检查特殊字符还是什么?我知道Fortran对输入很挑剔
编辑:在跟踪错误的代码序列下方
1) 主模块打开名称列表
write (unit=*,fmt='(a)') 'Reading namelist information'
call read_nl(trim(name_name))
read\nl
subroutine read_nl(namelist_name)
use ename_coms, only : nl & ! intent(inout)
, init_ename_vars ! ! subroutine
implicit none
!----- Arguments. ----------------------------------------------------------------------!
character(len=*), intent(in) :: namelist_name
!----- Local variables. ----------------------------------------------------------------!
logical :: fexists
!----- Name lists. ---------------------------------------------------------------------!
namelist /ED_NL/ nl
!---------------------------------------------------------------------------------------!
!----- Open the namelist file. ---------------------------------------------------------!
inquire (file=trim(namelist_name),exist=fexists)
if (.not. fexists) then
call fatal_error('The namelist file '//trim(namelist_name)//' is missing.' &
,'read_nl','ed_load_namelist.f90')
end if
!----- Initialise the name list with absurd, undefined values. -------------------------!
call init_ename_vars(nl)
!----- Read grid point and options information from the namelist. ----------------------!
open (unit=10, status='OLD', file=namelist_name)
read (unit=10, nml=ED_NL)
close(unit=10)
return
end subroutine read_nl
2) 然后是关于如何(哪些变量)从名称列表(输入)中读取的细节
我的模拟是'NOT_HISTORY'
:copy_nl('ALL_CASES')
被指定,因此它读取许多名称列表变量,包括变量'runtype'
,然后在if-else语句中使用
3) 然后copy\u nl('NOT\u HISTORY')
如下所示,这是定义nzg和nzs的地方
subroutine copy_nl(copy_type)
use grid_coms , only : time & ! intent(out)
, centlon & ! intent(out)
, centlat & ! intent(out)
, deltax & ! intent(out)
, deltay & ! intent(out)
, nnxp & ! intent(out)
, nnyp & ! intent(out)
, nstratx & ! intent(out)
, nstraty & ! intent(out)
, polelat & ! intent(out)
, polelon & ! intent(out)
, ngrids & ! intent(out)
, timmax & ! intent(out)
, time & ! intent(out)
, nzg & ! intent(out)
, nzs ! ! intent(out)
implicit none
!----- Arguments. ----------------------------------------------------------------------!
character(len=*), intent(in) :: copy_type
!----- Internal variables. -------------------------------------------------------------!
integer :: ifm
!---------------------------------------------------------------------------------------!
!---------------------------------------------------------------------------------------!
! Here we decide which variables we should copy based on the input variable. !
!---------------------------------------------------------------------------------------!
select case (trim(copy_type))
case ('NOT_HISTORY')
itimea = nl%itimea
idatea = nl%idatea
imontha = nl%imontha
iyeara = nl%iyeara
nzg = nl%nzg
nzs = nl%nzs
slz(1:nzgmax) = nl%slz(1:nzgmax)
current_time%year = iyeara
current_time%month = imontha
current_time%date = idatea
current_time%time = real(int(real(itimea) * 0.01)) * hr_sec &
+ (real(itimea) * 0.01 - real(int(real(itimea)*0.01))) &
* 100.0 * min_sec
time = 0.0d0
请注意,我没有将所有的模块和变量都放在use
中,否则这篇文章会非常长。
仅供参考,模块grid\u coms
指定变量的类型。参见下面的相关部分(整个模块是一个变量列表,没有其他内容)
4) 回到主模块,然后调用ed\u opspec\u grid
检查变量是否合理,这就是问题所在。同样地,使用use
时,变量在开始时就被初始化了,我在这里不讨论这个问题
subroutine ed_opspec_grid
!---------------------------------------------------------------------------------------!
! Check whether soil layers are reasonable, i.e, enough layers, sorted from the !
! deepest to the shallowest. !
!---------------------------------------------------------------------------------------!
if (nzg < 2) then
write (reason,'(a,1x,i4,a)') &
'Too few soil layers. Set it to at least 2. Your nzg is currently set to' &
,nzg,'...'
call opspec_fatal(reason,'opspec_grid')
ifaterr=ifaterr+1
elseif (nzg > nzgmax) then
write (reason,'(2(a,1x,i5,a))') &
'The number of soil layers cannot be greater than ',nzgmax,'.' &
,' Your nzg is currently set to',nzg,'.'
call opspec_fatal(reason,'opspec_grid')
ifaterr=ifaterr+1
end if
do k=1,nzg
if (slz(k) > -.001) then
write (reason,'(a,1x,i4,1x,a,1x,es14.7,a)') &
'Your soil level #',k,'is not enough below ground. It is currently set to' &
,slz(k),', make it deeper than -0.001...'
call opspec_fatal(reason,'opspec_grid')
ifaterr=ifaterr+1
end if
end do
do k=1,nzg-1
if (slz(k)-slz(k+1) > .001) then
write (reason,'(2(a,1x,i4,1x),a,2x,a,1x,es14.7,1x,a,1x,es14.7,a)') &
'Soil layers #',k,'and',k+1,'are not enough apart (i.e. > 0.001).' &
,'They are currently set as ',slz(k),'and',slz(k+1),'...'
call opspec_fatal(reason,'opspec_grid')
ifaterr=ifaterr+1
end if
end do
end subroutine ed_opspec_grid
子程序ed\u opspec\u网格
!---------------------------------------------------------------------------------------!
! 检查土层是否合理,即是否有足够的土层,从!
! 最深最浅!
!---------------------------------------------------------------------------------------!
如果(nzg<2),则
写入(原因,“(a,1x,i4,a)”&
“土层太少。将其设置为至少2。您的nzg当前设置为'&
,nzg,“…”
调用opspec\u fatal(原因为'opspec\u grid')
ifaterr=ifaterr+1
elseif(nzg>nzgmax)然后
写入(原因,“(2(a,1x,i5,a)))&
“土壤层的数量不能大于”,nzgmax“。”&
,'您的nzg当前设置为',nzg'。'
调用opspec\u fatal(原因为'opspec\u grid')
ifaterr=ifaterr+1
如果结束
do k=1,nzg
如果(slz(k)>-.001),那么
写入(原因,“(a,1x,i4,1x,a,1x,es14.7,a)”&
“你的土壤水平面”k“在地下是不够的。它当前设置为'&
,slz(k),',使其深度超过-0.001…'
调用opspec\u fatal(原因为'opspec\u grid')
ifaterr=ifaterr+1
如果结束
结束
Dok=1,nzg-1
如果(slz(k)-slz(k+1)>.001),那么
写入(原因,“(2(a,1x,i4,1x),a,2x,a,1x,es14.7,1x,a,1x,es14.7,a)”&
“土壤层#”,k,“和”,k+1,“距离不够(即>0.001)。”&
,“它们当前设置为',slz(k),'和',slz(k+1),'…”
调用opspec\u fatal(原因为'opspec\u grid')
ifaterr=ifaterr+1
如果结束
结束
结束子程序ed_opspec_网格
请注意,这不是此子例程中的第一次检查。在这一部分之前检查了其他变量(我省略了这些变量),但这是第一条错误消息。这让我觉得,可能部分输入是可读的,而有些是不可读的
最后,让我再次强调,这是一个非常大的项目,我真的无法显示所有的代码,这就是为什么我非常简单地提出了这个问题:Fortran是否有我可能遗漏的任何(文本)输入要求(特殊字符、返回,对于不同的系统可能不同?)。
另外,这段代码被很多研究人员在不同的平台上使用,所以我很怀疑代码本身是否有问题。。。(但如果我错了,请告诉我)。您没有为我们提供足够的数据来确定问题,因此我将使用姓名列表告诉您我遇到的两个问题:
a
中,名称列表b
,但代码先读取b
,则除非您倒带文件,否则它将找不到a
-999
,以便注意到它们的缺失。因此,请仔细检查数据文件是否正确。查找可能过早结束姓名列表的离群/
字符namelist/../…,…,…
规范module grid_coms
integer :: nzg ! Number of soil levels
integer :: nzs ! Number of snow/surface water levels
end module grid_coms
subroutine ed_opspec_grid
!---------------------------------------------------------------------------------------!
! Check whether soil layers are reasonable, i.e, enough layers, sorted from the !
! deepest to the shallowest. !
!---------------------------------------------------------------------------------------!
if (nzg < 2) then
write (reason,'(a,1x,i4,a)') &
'Too few soil layers. Set it to at least 2. Your nzg is currently set to' &
,nzg,'...'
call opspec_fatal(reason,'opspec_grid')
ifaterr=ifaterr+1
elseif (nzg > nzgmax) then
write (reason,'(2(a,1x,i5,a))') &
'The number of soil layers cannot be greater than ',nzgmax,'.' &
,' Your nzg is currently set to',nzg,'.'
call opspec_fatal(reason,'opspec_grid')
ifaterr=ifaterr+1
end if
do k=1,nzg
if (slz(k) > -.001) then
write (reason,'(a,1x,i4,1x,a,1x,es14.7,a)') &
'Your soil level #',k,'is not enough below ground. It is currently set to' &
,slz(k),', make it deeper than -0.001...'
call opspec_fatal(reason,'opspec_grid')
ifaterr=ifaterr+1
end if
end do
do k=1,nzg-1
if (slz(k)-slz(k+1) > .001) then
write (reason,'(2(a,1x,i4,1x),a,2x,a,1x,es14.7,1x,a,1x,es14.7,a)') &
'Soil layers #',k,'and',k+1,'are not enough apart (i.e. > 0.001).' &
,'They are currently set as ',slz(k),'and',slz(k+1),'...'
call opspec_fatal(reason,'opspec_grid')
ifaterr=ifaterr+1
end if
end do
end subroutine ed_opspec_grid