[llvm-dev] [LAI] Question about getDependences

[Dangeti Tharun kumar via llvm-dev]

A gentle reminder.

On Fri, Feb 19, 2021 at 12:27 AM Dangeti Tharun kumar <
cs15mtech11002 at iith.ac.in> wrote:

> Hi Florian,
>
> Thanks for the reply.
>
> We have tried a condition like this:
>
> *if (LAI.getMaxSafeDepDistBytes() == -1ULL && !LAI.canVectorizeMemory()) {*
> *  print("Unknown dependences in loop");*
> *  return false;*
> *}*
>
> Looks like *LAI.getMaxSafeDepDistBytes() *is *not always *capturing the
> distance for non-vectorizable loops.
>
> This example from TSVC, is not vectorizable but dependences are analyzable.
>
> for (int i = 1; i < LEN_1D - 1; i++) {
>       a[i] = b[i - 1] + c[i] * d[i];
>       b[i] = b[i + 1] - e[i] * d[i];
>  }
>
>
> * LAI.getMaxSafeDepDistBytes() is : (ULL)-1*
> * LAI.canVectroizeMemory is : false*
>
> Any more suggestions would be appreciated. :)
>
>
>
>
> On Thu, Feb 18, 2021 at 11:54 PM Florian Hahn <florian_hahn at apple.com>
> wrote:
>
>> Hi,
>>
>> On 18 Feb 2021, at 18:17, Dangeti Tharun kumar via llvm-dev <
>> llvm-dev at lists.llvm.org> wrote:
>>
>> Hi,
>>
>> I am using LAI->getDependences(), It returns a vector of Dependence
>> objects in a Loop.
>> If the loop is un-analyzable or parallel, an empty list is returned.
>>
>> *for example:*
>>
>> *Loop A:*
>> for (int i = 0; i < N; i ++) {
>>         a[i] = c[i] + d[i];
>>         b[i] = c[i] * d[i];
>> }
>> *Loop B:*
>> for (int i = 0; i < N; i++) {
>>      a[b[i]] = c[i] + d[i];
>>      c[i] = c[i  - 1] + 1;
>> }
>>
>> Loop A is parallel and Loop B has non-simple dependences(LAI bails out)
>> for both the cases we get an empty set of dependences.
>>
>> Is there an API that can tell if LAI was successful in dependence
>> computation?
>>
>>
>> I think canVectorizeMemory should do what you want (possibly in
>> combination with getMaxSafeDepDistBytes
>>
>> Cheers,
>> Florian
>>
>
>
> --
> Regards,
> DTharun
>


-- 
Regards,
DTharun

-- 


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