https://gcc.gnu.org/bugzilla/show_bug.cgi?id=30409
--- Comment #11 from anlauf at gcc dot gnu.org ---
(In reply to kargl from comment #10)
> (In reply to anlauf from comment #8)
> I'm not sure what you are worried about here. If one has
>
> do i = 1, n
> ... = expression1(..., 1/y)
> end do
>
> then this is equivalent to
>
> do i = 1, n
> tmp = 1 / y
> ... = expression1(..., tmp)
> end do
OK so far.
> which is equivalent to
>
> tmp = 1 / y
> do i = 1, n
> ... = expression1(..., tmp)
> end do
No. Strictly speaking, it is only equivalent to:
if (n > 0) tmp = 1 / y
do i = 1, n
... = expression1(..., tmp)
end do
