我想在Fortran代码中为派生类型实现用户定义的I / O过程。但是,这些过程中的write
语句无法在两个连续的write
语句之间生成新行。派生类型和程序定义如下。
模块:
module station_module
implicit none
character(8), parameter :: FmtFloat = '(5E15.7)'
type :: station
integer, private :: ns = 0
real, public, allocatable :: xloc(:), yloc(:), zloc(:)
contains
procedure, public :: import_station
procedure, public :: export_station
procedure, private :: read_station
generic, public :: read (formatted) => read_station
procedure, private :: write_station
generic, public :: write (formatted) => write_station
final :: destruct_station
end type station
interface station
module procedure new_station
end interface station
contains
function new_station(n) result(t)
implicit none
integer, intent(in) :: n
type(station) :: t
if (n > 0) then
allocate (t%zloc(n))
allocate (t%yloc(n))
allocate (t%xloc(n))
t%ns = n
end if
end function new_station
subroutine read_station(dtv, unit, iotype, vlist, iostat, iomsg)
implicit none
class(station), intent(inout) :: dtv
integer, intent(in) :: unit
character(*), intent(in) :: iotype
integer, intent(in) :: vlist(:)
integer, intent(out) :: iostat
character(*), intent(inout) :: iomsg
call dtv%import_station(unit)
iostat = 0
end subroutine read_station
subroutine import_station(this, unit)
implicit none
class(station), intent(inout) :: this
integer, intent(in) :: unit
character(256) :: header, footer
integer ns
read (unit, '(A)') header !> Header
read (unit, *) ns
if (ns > 0) then
if (allocated(this%zloc)) then
deallocate (this%zloc)
end if
allocate (this%zloc(ns))
read (unit, *) this%zloc
if (allocated(this%yloc)) then
deallocate (this%yloc)
end if
allocate (this%yloc(ns))
read (unit, *) this%yloc
if (allocated(this%xloc)) then
deallocate (this%xloc)
end if
allocate (this%xloc(ns))
read (unit, *) this%xloc
this%ns = ns
end if
read (unit, '(A)') footer !> Footer
end subroutine import_station
subroutine export_station(this, unit)
implicit none
class(station), intent(in) :: this
integer, intent(in) :: unit
write (unit, '(A)') ">STATION INFO"
write (unit, '(I6)') this%ns
write (unit, *) "Z:"
write (unit, FmtFloat) this%zloc
write (unit, *) "Y:"
write (unit, FmtFloat) this%yloc
write (unit, *) "X:"
write (unit, FmtFloat) this%xloc
write (unit, '(A)') ">END STATION"
end subroutine export_station
subroutine write_station(dtv, unit, iotype, vlist, iostat, iomsg)
implicit none
class(station), intent(in) :: dtv
integer, intent(in) :: unit
character(*), intent(in) :: iotype
integer, intent(in) :: vlist(:)
integer, intent(out) :: iostat
character(*), intent(inout) :: iomsg
call dtv%export_station(unit)
iostat = 0
end subroutine write_station
subroutine destruct_station(this)
implicit none
type(station), intent(inout) :: this
if (allocated(this%xloc)) then
deallocate (this%xloc)
end if
if (allocated(this%yloc)) then
deallocate (this%yloc)
end if
if (allocated(this%zloc)) then
deallocate (this%zloc)
end if
this%ns = 0
end subroutine destruct_station
end module station_module
我们可以看到用户定义的格式化的写语句只调用一个名为export_station
的常规子例程,我希望通过这两个方法得到相同的结果。
这是我的测试计划:
program Test
use station_module
implicit none
type(station) :: pt, pt1, pt2
pt = station(4)
write(*, *) pt
call pt%export_station(6)
end program Test
输出:
>STATION INFO 4Z: 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00
Y: 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00X: 0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00>END STATION
>STATION INFO
4
Z:
0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00
Y:
0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00
X:
0.0000000E+00 0.0000000E+00 0.0000000E+00 0.0000000E+00
>END STATION
常规子例程export_station
产生了我的期望。在两个write
语句之间生成新行,而派生类型的write
语句则不生成。
答案 0 :(得分:6)
此问题也在英特尔论坛上提出。我在那里回答。"用户定义的派生类型I / O都是非推进的(你不能改变它)。如果你想要换行,你必须明确地写它们(例如,使用/ format。)"
答案 1 :(得分:2)
这里有两类输出语句:父语句和子语句。第一种情况下的父输出语句是write (*,*) pt
。
当第一个是父项时,对export_station
到write_station
的调用会导致写语句出现子输出语句。当用户直接调用export_station
时,这些写语句本身就是父输出语句。
子数据传输语句和父数据传输语句之间的一个显着区别是父语句在数据传输之前和之后定位文件。也就是说,当write (unit,*) "Z:"
完成时,文件位于仅在传输语句为父级时才写入的记录之后。
因此,您会看到新的行:这只是放在书面记录之后。
子数据传输语句(不是在完成时定位文件)不会影响新行。
我目前无法访问测试机器,所以这部分是推测性的。您可以显式写入从new_line('')
返回的换行符,作为子转移语句输出的一部分。由于advance='no'
将在子语句中被忽略,您可以在两种情况下使用它,明确控制新行的写入位置,而不是依赖于当前存在的拆分记录方法。