Hi Jakub,

thanks for the comment & spec referral. I have now updated the patch –
and included the new C/Fortran testcase.

On 04.07.22 16:53, Jakub Jelinek via Fortran wrote:
> See OpenMP 5.2 [59:31-34]:
> A modifier that is an expression must neither lexically match the name of a simple modifier
> defined for the clause that is an OpenMP keyword nor modifier-name parenthesized-tokens,
> where modifier-name is the modifier-name of a complex modifier defined for the clause and
> parenthesized-tokens is a token sequence that starts with ( and ends with ).
>
> So, ref can't be step expression because it lexically matches the name of a
> simple modifier, so linear (var : ref) is equivalent to old style linear (ref (var):1)
> while e.g. linear (var : ref + 0) is equivalent to linear (var : step (ref + 0))
I see.
> You can certainly try to match "step ( %e ) )" or "step ( %e ) , " first,
> those would handle the case of valid complex modifier.
Done so + plus some more massage in order to support the following,
added as C/Fortran testcase:
> But, I think if there is
>    interface
>      integer function step (x, y, z)
>        integer :: x, y, z
>      end function step
>    end interface
> then
>    linear (v : step (x, y, z))
> should be rejected, not accepted as valid
>    linear (v : step (step (x, y, z)))
>
> I think I should add:
> int step (int x, int y, int z) { return x + y + z; }
>
> int
> foo (int x)
> {
>    int i;
>    #pragma omp parallel for linear (x : step (step (1, 2, 3)))
>    for (i = 0; i < 64; i++)
>      x += 6;
>    return x;
> }
>
> int
> bar (int x)
> {
>    int i;
>    #pragma omp parallel for linear (x : step (1, 2, 3))       /* { dg-error "expected" } */
>    for (i = 0; i < 64; i++)
>      x += 6;
>    return x;
> }
> as another testcase (where foo used to be invalid before and bar used to be
> valid).
Tobias
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