[PATCH] Loop Rerolling Pass

[Hal Finkel]

----- Original Message -----
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> On Oct 16, 2013, at 9:18 AM, Hal Finkel < hfinkel at anl.gov > wrote:
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> ----- Original Message -----
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> On 15 October 2013 22:11, Hal Finkel < hfinkel at anl.gov > wrote:
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> I made use of SCEV everywhere that I could (I think). SCEV is used to
> analyze the induction variables, and then at the end to help with
> the rewriting. I don't think that I can use SCEV everywhere,
> however. For one thing, I need to check for equivalence of
> instructions, with certain substitutions, for instructions (like
> function calls) that SCEV does not interpret.
> 
> 
> Hi Hal,
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> This is probably my lack of understanding of all that SCEV does than
> anything else.
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> My comment was to the fact that you seem to be investigating specific
> cases (multiply, adding, increment size near line 240), which SCEV
> could get that as an expression, and possibly making it slightly
> easier to work with. I'll let other SCEV/LV experts to chime in,
> because I basically don't know what I'm talking about, here. ;)
> 
> Okay, I see what you mean. The code in this block:
> if (Inc == 1) {
> // This is a special case: here we're looking for all uses (except
> for
> // the increment) to be multiplied by a common factor. The increment
> must
> // be by one.
> if (I->first->getNumUses() != 2)
> continue;
> 
> This code does not use SCEV because, IMHO, there is no need. It is
> looking for a very particular instruction pattern where the
> induction variable has only two uses: one which increments it by one
> (SCEV has already been used to determine that the increment is 1),
> and the other is a multiply by a small constant. It is to catch
> cases like this:
> 
> for (int i = 0; i < 500; ++i) {
> foo(3*i);
> foo(3*i+1);
> foo(3*i+2);
> }
> 
> And so, aside from the increment, all uses of the IV are via the
> multiply. If we find this pattern, then instead of attempting to
> classify all IV uses as functions of i, i+1, i+2, ... we attempt to
> classify all uses of the multiplied IV that way.
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> I think the more general SCEV-based way to do this would be to
> recursively walk the def-use chains starting at phis, looking past
> simple arithmetic until reaching an IV users (see
> IVUsers::AddUsersIfInteresting). Then you group the users by their
> IV operand's SCEV expression. If the SCEVs advance by the same
> constant, then you have your unrolled iterations and it doesn't
> matter how the induction variable was computed.
> LSRInstance::CollectChains does something similar.

Thanks! Collecting all IV users may be overkill here, but this is something that I should play with.

While I have your attention (hopefully), why does SCEV not have a signed division representation? I suspect that it why SCEV won't give be a backedge-taken count for a loop like:

  for (int i = 0; i < n; i += 5) {
    ...
  }

Thanks again,
Hal

> 
> 
> -Andy
> 

-- 
Hal Finkel
Assistant Computational Scientist
Leadership Computing Facility
Argonne National Laboratory