Previously SimplifyCFG used getSetSize which returns an APInt that is 1 bit wider than the ConstantRange's bit width. In the reasonably common case that the ConstantRange is 64-bits wide, this requires returning a 65-bit APInt. APInt's can only store 64-bits without a memory allocation so this is inefficient.
The new method takes the 8 as an input and tells if the range contains more than that many elements without requiring any wider math.
git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@302385
91177308-0d34-0410-b5e6-
96231b3b80d8
/// Compare set size of this range with the range CR.
bool isSizeStrictlySmallerThan(const ConstantRange &CR) const;
+ // Compare set size of this range with Value.
+ bool isSizeLargerThan(uint64_t MaxSize) const;
+
/// Return the largest unsigned value contained in the ConstantRange.
APInt getUnsignedMax() const;
return (Upper - Lower).ult(Other.Upper - Other.Lower);
}
+bool
+ConstantRange::isSizeLargerThan(uint64_t MaxSize) const {
+ assert(MaxSize && "MaxSize can't be 0.");
+ // If this a full set, we need special handling to avoid needing an extra bit
+ // to represent the size.
+ if (isFullSet())
+ return APInt::getMaxValue(getBitWidth()).ugt(MaxSize - 1);
+
+ return (Upper - Lower).ugt(MaxSize);
+}
+
APInt ConstantRange::getUnsignedMax() const {
if (isFullSet() || isWrappedSet())
return APInt::getMaxValue(getBitWidth());
Span = Span.inverse();
// If there are a ton of values, we don't want to make a ginormous switch.
- if (Span.getSetSize().ugt(8) || Span.isEmptySet()) {
+ if (Span.isSizeLargerThan(8) || Span.isEmptySet()) {
return false;
}