Does greater operator ">" satisfy strict weak ordering?
Even if the standard refers to "less than" for arbitrary Compare functions, that only implies "less than" in the context of the ordering.
If I define an ordering by comparison function [](int a, int b) { return a > b; }, then an element is "less than" another in this ordering if its integer value is greater. That's because the ordering I've created is an ordering of the integers in reverse order. You shouldn't read < as "less than" in orderings. You should read it as "comes before".
Whenever x < y is a strict weak ordering then x > y is also a strict weak ordering, just with the reverse order.
Does greater operator “>” satisfy strict weak ordering?
The mathematical strict greater than relation is a strict weak ordering.
As for the operator in C++ langauge: For all integers types: Yes. In general: No, but in most cases yes. Same applies to strict less than operator.
As for the confusing quote, "is less than" in that context intends to convey that means that the the end result of the sort operation is a non-decreasing sequence i.e. objects are "less" or equal to objects after them. If std::greater is used as comparison object, then greater values are "lesser" in order.
This may be confusing, but is not intended to exclude strict greater than operator.
what is the case where > doesn't satisfy strict weak ordering?
Some examples:
- Overloaded operators that don't satisfy the properties.
>operator on pointers that do not point to the same array has unspecified result.>does not satisfy irreflexivity requirement for floating point types in IEEE-754 representation unless NaNs are excluded from the domain.