Fortran睡眠

Question

有誰知道在Fortran中給定毫秒級的睡眠方法嗎？我不想使用非便攜式系統調用，因此Fortran或C庫固有的任何內容都將是首選。

Answer 1

使用的Fortran ISO C綁定使用C庫睡眠以秒爲單位睡覺：

module Fortran_Sleep 

    use, intrinsic :: iso_c_binding, only: c_int 

    implicit none 

    interface 

     ! should be unsigned int ... not available in Fortran 
     ! OK until highest bit gets set. 
     function FortSleep (seconds) bind (C, name="sleep") 
      import 
      integer (c_int) :: FortSleep 
      integer (c_int), intent (in), VALUE :: seconds 
     end function FortSleep 

    end interface 

end module Fortran_Sleep 


program test_Fortran_Sleep 

    use, intrinsic :: iso_c_binding, only: c_int 

    use Fortran_Sleep 

    implicit none 

    integer (c_int) :: wait_sec, how_long 

    write (*, '("Input sleep time: ")', advance='no') 
    read (*, *) wait_sec 
    how_long = FortSleep (wait_sec) 

    write (*, *) how_long 

    stop 

end program test_Fortran_Sleep 

Answer 2

您可以使用Fortran標準的內在功能做到這一點無C綁定：

program sleep 
!=============================================================================== 
implicit none 
character(len=100) :: arg ! input argument character string 
integer,dimension(8) :: t ! arguments for date_and_time 
integer :: s1,s2,ms1,ms2 ! start and end times [ms] 
real :: dt    ! desired sleep interval [ms] 
!=============================================================================== 
! Get start time: 
call date_and_time(values=t) 
ms1=(t(5)*3600+t(6)*60+t(7))*1000+t(8) 

! Get the command argument, e.g. sleep time in milliseconds: 
call get_command_argument(number=1,value=arg) 
read(unit=arg,fmt=*)dt 

do ! check time: 
    call date_and_time(values=t) 
    ms2=(t(5)*3600+t(6)*60+t(7))*1000+t(8) 
    if(ms2-ms1>=dt)exit 
enddo 
!=============================================================================== 
endprogram sleep 

假設可執行文件是SLP：

~$ time slp 1234 

real  0m1.237s 
user  0m1.233s 
sys   0m0.003s 

一個特例加入這一計劃如果y你擔心它會在午夜時分左右:)