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From: "jakub at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>
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Subject: [Bug fortran/103662] [12 Regression] TBAA problem in Fortran FE
 triggering in gfortran.dg/unlimited_polymorphic_3.f03
Date: Tue, 19 Apr 2022 14:07:07 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D103662

--- Comment #15 from Jakub Jelinek <jakub at gcc dot gnu.org> ---
Putting a breakpoint on new_alias_set, I see that
RECORD_TYPE sq gets alias set 1
RECORD_TYPE s gets alias set 9
RECORD_TYPE t gets alias set 10

Those are
  type sq
    sequence
    integer :: i
  end type sq
and
  type, bind(c) :: s
    integer (c_int) :: k
  end type s
  type t
    sequence
    integer :: k
  end type t
in the testcase.  The testcase stores through the sq type and reads through
pointer to s and pointer to t type.  As the alias sets are different, for t=
he
middle-end the testcase is invalid.
Now, the question is what is the Fortran unlimited polymorphic semantics, if
one can store through one type and read through a different type which just=
 has
the same layout, or if it also has to use the same name etc., or if it is e=
ven
acceptable to store through say a type with a pair of integers and read thr=
ough
a type with double precision etc.
GCC has types that can alias anything, so if the semantics is like that,
unlimited polymorphic middle-end pointers perhaps should use that.
Or the testcase is invalid and has to use type sq in foo as well.  Or we ne=
ed
to ensure sq and s/t above have the same alias set.=