procedure pointer issue

procedure pointer issue

Hello, I am having trouble with calling a procedure via pointer. I am not sure if this is my misunderstanding of the fortran standard, or a compiler problem. In the piece of code shown below, a call to a procedure pointer does not produce the expected results. Arrays 'n1' and 'n2', previously allocated with a certain size ('hmuch') when passed to a pointer 'proc', give an 'ubound' value of 0. If passed to 'a_1' , or 'a_2' directly (so, 'call a_1(n1,n2)' instead of call 'proc(n1,n2)'), ubound returns the correct sizes of the arrays.

The way I see it this is one of the following: 1) My code is correct, this is a compiler bug 2) My code has an intricate error, but compiler compiles blissfully, instead of signaling a dangerous omission which has apparently something to do with the interface information available in the main program. 3) This is a daft error on my side, the one i wouldn't make have I stopped coding hours ago and resumed after recharging my mana. Can somebody help me figure out what is actually going wrong?

The version of the compiler is: ifort (IFORT) 11.1 20091012


module procs

 implicit none


   subroutine a_1(x,y)

      complex,dimension(:),intent(in) :: x,y !I should be able to pass allocatable arrays to this subroutine, right?

      print *, 'x', ubound(x)

      print *, 'y', ubound(y)

   end subroutine a_1 
   subroutine a_2(x,y)

     complex,dimension(:),intent(in) :: x,y

     print *, 'xx', ubound(x) print *, 'yy', ubound(y)

   end subroutine a_2

end module procs 
program bullgrog

  use procs

  implicit none

  procedure(), pointer :: proc !the procedure pointer used further below to call a_1 or a_2, depending on the value of 'sel'

  integer :: sel,hmuch

  complex,dimension(:),allocatable :: n1,n2

  if (sel .eq. 1) then

    proc => a_1          !so 'proc' will point to 'a_1'


    proc => a_2

  end if 

  print *, 'n1', ubound(n1)     !30

  print *, 'n2', ubound(n2)     !30 
  call proc(n1,n2)                   !THE PROBLEM --> I would have thought the result is 'x 30 n y 30', but, no, it is 'x 0 n y 0' !! Huh??

  call a_2(n1,n2)                    ! a direct call returns 'xx 30 n yy 30' like i thought

end program bullgrog

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