================
@@ -1520,15 +1520,72 @@ ConstantRange ConstantRange::binaryNot() const {
return ConstantRange(APInt::getAllOnes(getBitWidth())).sub(*this);
}
+/// Estimate the 'bit-masked AND' operation's lower bound.
+///
+/// E.g., given two ranges as follows (single quotes are separators and
+/// have no meaning here),
+///
+/// LHS = [10'00101'1, ; LLo
+/// 10'10000'0] ; LHi
+/// RHS = [10'11111'0, ; RLo
+/// 10'11111'1] ; RHi
+///
+/// we know that the higher 2 bits of the result is always 10; and we also
+/// notice that RHS[1:6] are always 1, so the result[1:6] cannot be less than
+/// LHS[1:6] (i.e., 00101). Thus, the lower bound is 10'00101'0.
+///
+/// The algorithm is as follows,
+/// 1. we first calculate a mask to find the higher common bits by
+/// Mask = ~((LLo ^ LHi) | (LLo ^ LHi) | (LLo ^ RLo));
+/// Mask = clear all non-leading-ones bits in Mask;
+/// in the example, the Mask is set to 11'00000'0;
+/// 2. calculate a new mask by setting all common leading bits to 1 in RHS, and
+/// keeping the longest leading ones (i.e., 11'11111'0 in the example).
+/// 3. return (LLo & new mask) as the lower bound;
+/// 4. repeat the step 2 and 3 with LHS and RHS swapped, and update the lower
+/// bound with the larger one.
+static APInt estimateBitMaskedAndLowerBound(const ConstantRange &LHS,
----------------
zsrkmyn wrote:
> I happened to analyze the problem in Oct and made a comment to
> https://stackoverflow.com/questions/2620388/bitwise-interval-arithmetic
>
> Do the two algorithms return the same result? If yes, my code seems simpler?
Aha, I saw your answer on SO before, but I don't think they are same. E.g.,
given
```
a.lo = 0b110101;
a.hi = 0b111000; // inclusive
b.lo = 0b111111;
b.hi = 0b111111;
```
your algorithm gives the lower bound as `0b110000`, while mine gives `0b110101`.
https://github.com/llvm/llvm-project/pull/120352
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