time_sub()

subroutine time_sub	(	integer, dimension(7)	date1,
		integer, dimension(7)	date2,
		integer, dimension(7)	date3
	)

Definition at line 31 of file time_sub.f.

References time_copy(), and time_norm().

c
c date3=abs(date1-date2)
c 
c This routine will only return the absolute value of the result.
c We do not handle signs. Therefore in any case the smaller time
c value will be subtracted from the larger one.
c
c relative - relative -> relative   case 1
c absolute - relative -> absolute   case 2
c absolute - absolute -> relative   case 3
c
c NOTICE: This version of time_sub disregards leap-seconds!
c
c input:
c   date1:  date record
c   date2:  date record
c output:
c   date3:  fully qualified and regularized absolute difference between
c           input date values (maybe rælative or absolute - see above)
c
c last change: V2.00 (05/08/2000)
c
      integer date1(7), date2(7), date3(7)
cE
      integer i, larger(7), smaller(7), case, time_compare
      logical time_isleapyear
c
      call time_norm(date1)
      call time_norm(date2)
c find out case and set larger and smaller
      if ((date1(1).eq.0).or.(date2(1).eq.0)) then
        if (date1(1).eq.date2(1)) then
          case=1
        else
          case=2
          if (date1(1).eq.0) then
            call time_copy(date2, larger)
            call time_copy(date1, smaller)
          else
            call time_copy(date1, larger)
            call time_copy(date2, smaller)
          endif
        endif
      else
        case=3
      endif
c 
      if (case.ne.2) then
        if (time_compare(date1, date2).gt.0) then
          call time_copy(date1, larger)
          call time_copy(date2, smaller)
        else
          call time_copy(date2, larger)
          call time_copy(date1, smaller)
        endif
      endif
c handle year boundaries of absolute times
      if (case.eq.3) then
    1   if (larger(1).eq.smaller(1)) goto 2        
          larger(1)=larger(1)-1
          larger(2)=larger(2)+365
          if (time_isleapyear(larger(1))) larger(2)=larger(2)+1
          goto 1
    2   continue
      endif
c build difference
      do i=1,7
        date3(i)=larger(i)-smaller(i)
      enddo
      call time_norm(date3)
      return

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