Questions on "forrtl: error (65): floating invalid"

Question

I've been running a huge Fortran code with Intel compiler version 13.1.3.192 under the debug mode (with -O0 -g -traceback -fpe3 flags being turned on). It gave me the following output message:

... ...
forrtl: warning (402): fort: (1): In call to MPI_ALLGATHER, an array temporary was created for argument #1

forrtl: error (65): floating invalid
Image              PC                Routine            Line        Source
arts               00000000016521D9  pentadiagonal_ser         385  pentadiagonal.f90
arts               0000000001644862  pentadiagonal_             62  pentadiagonal.f90
arts               00000000004DF167  implicit_solve_x_        1201  implicit.f90
arts               0000000000538079  scalar_bquick_inv         383  scalar_bquick.f90
arts               00000000004EFEAC  scalar_step_              190  scalar.f90
arts               0000000000401744  simulation_run_           182  simulation.f90
arts               0000000000401271  MAIN__                     10  main.f90
arts               0000000000400FAC  Unknown               Unknown  Unknown
arts               000000000420E444  Unknown               Unknown  Unknown
arts               0000000000400E81  Unknown               Unknown  Unknown

and the source of the error comes from the subroutine pentadiagonal_serial, which is to solve a penta-diagonal matrix:

subroutine pentadiagonal_serial(A,B,C,D,E,R,n,lot)
  use precision
  implicit none

  integer, intent(in) :: n,lot
  real(WP), dimension(lot,n) :: A     ! LOWER-2
  real(WP), dimension(lot,n) :: B     ! LOWER-1
  real(WP), dimension(lot,n) :: C     ! DIAGONAL
  real(WP), dimension(lot,n) :: D     ! UPPER+1
  real(WP), dimension(lot,n) :: E     ! UPPER+2
  real(WP), dimension(lot,n) :: R     ! RHS - RESULT
  real(WP), dimension(lot) :: const
  integer :: i

  if (n .eq. 1) then
    ! Solve 1x1 system
    R(:,1) = R(:,1)/C(:,1)
    return
  else if (n .eq. 2) then
    ! Solve 2x2 system
    const(:) = B(:,2)/C(:,1)
    C(:,2) = C(:,2) - D(:,1)*const(:)
    R(:,2) = R(:,2) - R(:,1)*const(:)
    R(:,2) = R(:,2)/C(:,2)
    R(:,1) = (R(:,1) - D(:,1)*R(:,2))/C(:,1)
    return
 end if

 ! Forward elimination
 do i=1,n-2
     ! Eliminate A(2,i+1)
     const(:) = B(:,i+1)/(C(:,i)+tiny(1.0_WP))
     C(:,i+1) = C(:,i+1) - D(:,i)*const(:)
     D(:,i+1) = D(:,i+1) - E(:,i)*const(:)
     R(:,i+1) = R(:,i+1) - R(:,i)*const(:)

in which the line

  const(:) = B(:,i+1)/(C(:,i)+tiny(1.0_WP))

causes the error. I tried to print out the values of const(:) and find that there did exist Infinity values. However, I cannot understand why it can generate infinities. As far as I see, to avoid zero denominator, the tiny(1.0_WP) is added to C(:,i) and now it is nearly impossible for the denominator to be zero... I also checked when this subroutine is being called, everything is initialized or given a value after declaration. So I couldn't figure out where could go wrong.

Answer 1

(Answers in the comments. See Question with no answers, but issue solved in the comments (or extended in chat). There is a lot of chat in the comments and its hard to extract the actual answer, but the OP indicates it is solved. )

@SethMMorton wrote:

If any of the values of C are around -tiny then you still can end up with zero in the denominator.

Did you print out the values of C? Do small values correlate to the infinities in const?

@Alexander Vogt wrote:

Well, adding something to infinity results in a floating point exception... And I'm not sure ifort can detect that a division by infinity is zero, I guess that is another floating point exception! You'd better make sure that no infinite values occur inside C.

@Stefan wrote:

Workaround: What about the routine ?gbsv from MKL/Lapack? Then, one last hint: Try to compile with gfortran. Different compilers create different compile-time/run-time warnings and/or errors and maybe gfortran detects your problem earlier.