It's 4 5-bit values, not 5 4-bit values. 5-bits is enough to enumerate each bit in a 32-bit bitfield. This means for any application that might set 4 or fewer f

It's 4 5-bit values, not 5 4-bit values. 5-bits is enough to enumerate each bit in a 32-bit bitfield. This means for any application that might set 4 or fewer flags out of a possible 32 need only use a 16-bit bitfield. As I said in another comment I just tested this with a 32-bit hash array mapped trie, and it would result in a significant reduction in the storage overhead of the trie (which is already more efficient than typical hash tables for map like data structures).