Is there a particular reason the less than operator is used as default comparison for max pq?

Question

Why couldn't the greater operator be used as the comparison? Is there a particular advantage in using less than?

All the other STL facilities use less-than. Consequently, a lot of classes are written that implement `operator<` but not necessarily `operator>`. If `priority_queue` weren't consistent, it won't be usable with such classes without extra work. So basically, it's to minimize the size of the interface that the user needs to provide in order to use the library. — Igor Tandetnik, Nov 25 '18 at 17:00
There is no advantage, technically. It's just for consistency. Besides, the minimum requirement for sorting is defined with by some ordering relationship, and the STL chose some 'form of` operator <` — WhiZTiM, Nov 25 '18 at 17:01
If you want to be equal opportunity for all operators, check out the boost class template boost::less_than_comparable https://stackoverflow.com/questions/4589353/what-is-the-minimum-set-of-operators-i-need-to-overload/4589464?s=2|37.0776#4589464 — natersoz, Nov 25 '18 at 17:06
As long as the operators implement/impose a strict weak ordering, `std::less` and `std::greater` could each be the default. But, you have to pick one, and less seems more natural most of the time, so it's a sane default. You can always override it if you need to. — Jesper Juhl, Nov 25 '18 at 17:14

score 4 · Answer 1 · answered Nov 25 '18 at 17:02

There are two aspects to your question:

Firstly, that's technically not correct: Most algorithms (not sure which one exactly you mean) use std::less as a default. However, that default can be overridden. The important point is the behaviour of the comparison, i.e. that it imposes a "strict-weak ordering", a term you will find mentioned in many texts on the topic and (I believe) even the standard itself.
Secondly, doing things in ascending order is often "natural" and it is the default throughout the C++ standard library.

1 Answers1