(j3.2006) Did we intend to prohibit this?
Clune, Thomas L. GSFC-6101
Fri Mar 10 16:47:01 EST 2017
> On Mar 10, 2017, at 2:09 PM, Van Snyder <Van.Snyder at jpl.nasa.gov> wrote:
> On Fri, 2017-03-10 at 16:13 +0900, Cohen Malcolm wrote:
>>> So how does it get deallocated?
>> "it" is not a variable, so does not get deallocated (in the Fortran
>> language sense of "deallocated").
>> You know what the function result variable is, and it is not the value
>> that is returned by the function, nor has it ever been. The result
>> variable is what gets deallocated.
> So what does 22.214.171.124p4 mean?
> "If an executable construct references a function whose result
> is allocatable or has an allocatable subobject, and the function
> reference is executed, an allocatable result and any allocated
> allocatable subobject of the result is deallocated after
> execution of the innermost executable construct containing the
> If the result variable is in fact what gets deallocated, even though the
> word "variable" does not appear in 126.96.36.199p4, then the result variable
> of F(X) in
> call move_alloc ( f(x), v )
> IS a variable, right? So what's the problem with executing it?
The result variable of F(X) is not in the expression you provide. Outside of F all you can ?see" is the result _value_.
I think that what you probably want in this situation is:
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