ScalarEvolution: Compute exit counts for loops with a power-of-2 step.

If we have a loop of the form
for (unsigned n = 0; n != (k & -32); n += 32) {}
then we know that n is always divisible by 32 and the loop must
terminate. Even if we have a condition where the loop counter will
overflow it'll always hold this invariant.

PR19183. Our loop vectorizer creates this pattern and it's also
occasionally formed by loop counters derived from pointers.

llvm-svn: 204728

1 files changed, 10 insertions, 0 deletions

diff --git a/llvm/lib/Analysis/ScalarEvolution.cpp b/llvm/lib/Analysis/ScalarEvolution.cpp
index 39bf95b92d9..08de6213e22 100644
--- a/llvm/lib/Analysis/ScalarEvolution.cpp
+++ b/llvm/lib/Analysis/ScalarEvolution.cpp
@@ -5744,6 +5744,16 @@ ScalarEvolution::HowFarToZero(const SCEV *V, const Loop *L, bool IsSubExpr) {
       getUDivExpr(Distance, CountDown ? getNegativeSCEV(Step) : Step);
     return ExitLimit(Exact, Exact, /*MustExit=*/false);
   }
+
+  // If Step is a power of two that evenly divides Start we know that the loop
+  // will always terminate.  Start may not be a constant so we just have the
+  // number of trailing zeros available.  This is safe even in presence of
+  // overflow as the recurrence will overflow to exactly 0.
+  const APInt &StepV = StepC->getValue()->getValue();
+  if (StepV.isPowerOf2() &&
+      GetMinTrailingZeros(getNegativeSCEV(Start)) >= StepV.countTrailingZeros())
+    return getUDivExactExpr(Distance, CountDown ? getNegativeSCEV(Step) : Step);
+
   // Then, try to solve the above equation provided that Start is constant.
   if (const SCEVConstant *StartC = dyn_cast<SCEVConstant>(Start))
     return SolveLinEquationWithOverflow(StepC->getValue()->getValue(),