[J3] Self-assignment of allocatable component

[Van Snyder]

The answer to this question needs to be consistent with the answer to
what happens with
call move_alloc ( a, a )
There was an interp about this.
On Mon, 2021-07-26 at 15:15 -0400, Daniel C Chen via J3 wrote:
> Is the following code standard conforming?
> type dt
>   integer, allocatable :: i
> end type
> 
> type(dt), pointer :: p
> type(dt), target :: t
> 
> allocate(t%i)
> t%i = 4
> p => t
> p = p       !! self-assignment
> 
> end
> From 21-007r1 [p172: 23-30]
> "For a noncoarray allocatable component the following sequence of
> operations is applied.
> (1) If the component of the variable is allocated, it is deallocated.
> (2) If the component of the value of expr is allocated, the
> corresponding component of the variable is
> allocated with the same dynamic type and type parameters as the
> component of the value of expr.
> If it is an array, it is allocated with the same bounds. The value of
> the component of the value of
> expr is then assigned to the corresponding component of the variable
> using defined assignment if the
> declared type of the component has a type-bound defined assignment
> consistent with the component,
> and intrinsic assignment for the dynamic type of that component
> otherwise.
> "
> It seems it is invalid because according to "(1)",  the allocatable
> component "i" is deallocated. However, Note 8 seems suggest it is
> still a conforming code as:
> "NOTE 8
> If an allocatable component of expr is unallocated, the corresponding
> component of the variable has an
> allocation status of unallocated after execution of the assignment."
> 
> Apprently p%i is unallocated after the self-assignment. Is this code
> still standard conforming?
> 
> Thanks,
> 
> Daniel Chen
> 
> XL Fortran Development, Fortran Standard Representative
> IBM Toronto Software Lab
> Phone: 905-413-3056   
> Tie: 969-3056   
> Email: cdchen at ca.ibm.com
> 
> 
> 
> 
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