[Bug fortran/11251] New: g77 complex number program produces wrong code when optimising

["jfb at npl dot co dot uk" <gcc-bugzilla at gcc dot gnu dot org>]

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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=11251

           Summary: g77 complex number program produces wrong code when
                    optimising
           Product: gcc
           Version: 3.2.2
            Status: UNCONFIRMED
          Severity: normal
          Priority: P2
         Component: fortran
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: jfb@npl.co.uk
                CC: gcc-bugs@gcc.gnu.org

The following program (complete) works correctly when I use a simple compile

g77 -o test.exe test.f

It produces the answer

(-8.,0.) (-10.,0.) (4.,0.)
(-38.,0.) (20.,0.) (-26.,0.)

However, when I compile the same program with optimisation

g77 -O -o test.exe test.f

It gives the wrong answer:

(-8.,0.) (-10.,0.) (4.,0.)
(-38.,0.) (110.,0.) (-86.,0.)  <<< this is wrong!

However, the -O2, -O3 and -O4 flags give the right answer!

Program listing follows:
-------------------------



      program test

c ----------------------------------------------------------------------
c     This program demonstrates a bug in g77. When compiled with
c     optimiser flag -O, it produces incorrect code and the second
c     result c is wrong. -O2,3 and 4 work ok! (and no optimisation).
c ----------------------------------------------------------------------

      complex a(3),b(3),c(3)
      
      a(1)=1
      a(2)=2
      a(3)=7

      b(1)=-1
      b(2)=2
      b(3)=3

      call cross(c,a,b)

      write (*,*) c

      call cross(c,c,b)

      write (*,*) c

      end

c ######################################################################

      subroutine cross(c,a,b)

c ----------------------------------------------------------------------
c     complex cross product
c     c=||a1 a2 a3||
c       ||b1 b2 b3||
c ----------------------------------------------------------------------

      complex a(3),b(3),c(3),r(3)

      r(1)=a(2)*b(3)-a(3)*b(2)
      r(2)=a(3)*b(1)-a(1)*b(3)
      r(3)=a(1)*b(2)-a(2)*b(1)

      c(1)=r(1)
      c(2)=r(2)
      c(3)=r(3)

      end