Results for “Std.BitSet.BitSet”
18 matching declarations
Use module:Std.List, kind:fn, or is:exact. Put a Pudu type after ::.
- Std.BitSet.BitSet
Sparse blocks of 64 IDs. Construct through this module to preserve block invariants.
- Std.BitSet.blocks
&Std.BitSet.BitSet -> Map[UInt64, UInt64]The words the set is stored as, keyed by where each begins.
- Std.BitSet.contains
&Std.BitSet.BitSet -> UInt64 -> BoolWhether a number is in the set.
- Std.BitSet.difference
&Std.BitSet.BitSet -> &Std.BitSet.BitSet -> Std.BitSet.BitSetEvery number in the first set and not the second.
- Std.BitSet.empty
Std.BitSet.BitSetA set holding nothing.
- Std.BitSet.fromArray
&Array[UInt64] -> Std.BitSet.BitSetA set holding each of these numbers.
- Std.BitSet.fromBlocks
&Map[UInt64, UInt64] -> Result[Std.BitSet.BitSet, UInt64]Validate block addresses and omit empty blocks.
- Std.BitSet.insert
&Std.BitSet.BitSet -> UInt64 -> Std.BitSet.BitSetThe set with a number added. Adding one already there changes nothing.
- Std.BitSet.intersection
&Std.BitSet.BitSet -> &Std.BitSet.BitSet -> Std.BitSet.BitSetEvery number in both sets.
- Std.BitSet.isDisjointFrom
&Std.BitSet.BitSet -> &Std.BitSet.BitSet -> BoolWhether the two sets share no number.
- Std.BitSet.isEmpty
&Std.BitSet.BitSet -> BoolWhether the set holds nothing.
- Std.BitSet.isSubsetOf
&Std.BitSet.BitSet -> &Std.BitSet.BitSet -> BoolWhether every number in the first set is in the second.
- Std.BitSet.remove
&Std.BitSet.BitSet -> UInt64 -> Std.BitSet.BitSetThe set with a number taken out, if it was there.
- Std.BitSet.singleton
UInt64 -> Std.BitSet.BitSetA set holding one number.
- Std.BitSet.size
&Std.BitSet.BitSet -> UInt128UInt128 can count even the entire UInt64 ID domain.
- Std.BitSet.symmetricDifference
&Std.BitSet.BitSet -> &Std.BitSet.BitSet -> Std.BitSet.BitSetEvery number in one set but not the other.
- Std.BitSet.toArray
&Std.BitSet.BitSet -> Array[UInt64]Materialize members in ascending order.
- Std.BitSet.union
&Std.BitSet.BitSet -> &Std.BitSet.BitSet -> Std.BitSet.BitSetEvery number in either set.
