ordered_set: operator<: Fix for a case of strict inclusion. #3431
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When all elements compared so far in sets
x
andy
have been equivalentand one of the sets (
x
) ends, but the other one (y
) does not end(i.e.
x
is a strict prefix ofy
),we used to get
y < x
, but with this changeset we are going to getx < y
.Example (in pseudocode) (with case-insensitive comparison):
we used to get
{'A', 'b'} > {'a', 'B', 'c'}
,now fixed to
{'A', 'b'} < {'a', 'B', 'c'}
.This is similar to how
std::lexicographical_compare
behaves.