[gomp4] Some progress on #pragma omp simd

[Aldy Hernandez]

On 04/24/13 01:01, Jakub Jelinek wrote:
> On Tue, Apr 23, 2013 at 09:32:29PM +0000, Iyer, Balaji V wrote:
>> 	My apologies if the documentation did not explain this correctly. It
>> was written by compiler developers and not language developers.  #pragma
>> simd is the guarantee the user gives the compiler that the inter-iteration
>> dependencies do not matter.  So, if the user omits the vectorlength the
>> clause then the compiler can, in effect, choose N, where N is the number
>> of loop iterations.
>
> The documentation doesn't suggest that.  Anyway, so
> #pragma simd
> should be equivalent to
> #pragma omp simd
> wrt. inter-iteration dependencies, and
> #pragma simd vectorlength(a, b, c)
> to
> #pragma omp simd safelen(max (a, b, c))
> ?  If so, then the FE could emit OMP_SIMD for #pragma simd, and if
> vectorlength is present, add OMP_CLAUSE_SAFELEN with the maximum of the
> values in all vectorlength clauses, and keep the vectorlength clauses around
> too as CILK_CLAUSE_VECTORLENGTH as hints to the vectorizer?

Well, it looks like things are bit simpler than expected.

Multiple vectorlength clauses are being deprecated or eliminated in the 
upcoming spec.  So it looks like vectorlength is the same thing as the 
safelen clause.

If you agree then I can get rid of OMP_CLAUSE_CILK_VECTORLENGTH and just 
emit an OMP_CLAUSE_SAFELEN.

Agreed?

>
> Also, Aldy said that #pragma simd loops allow != condition, how do you
> compute number of iterations in that case if the increment isn't constant?
> As conditional depending on whether increment is positive or negative?
> != condition isn't allowed in OpenMP, so there it is always obvious which
> direction it should iterate, and the expansion code will assume if it sees
> NE_EXPR that it is just folded border test (comparison with maximum or
> minimum value).

I verified with the Cilk Plus folks, and the number of iterations is 
calculated with a conditional.  So to evaluate something like this:

// incr = -1
for (i=N; i != limit; i += incr)
   [body]

We would generate something like this:

if (incr > 0) count =  (limit - N) / incr;
else count = (N - limit) / -incr;
for (i=N; count > 0; --count)
   [body]