Re: Is a std::map<> ordered?

Rolf Magnus <>
Fri, 21 Nov 2008 12:34:27 +0100
Pete Becker wrote:

On 2008-11-20 14:40:19 -0500, said:

On Nov 20, 7:48 am, Juha Nieminen <nos...@thanks.invalid> wrote:

joseph cook wrote:

A map is always sorted using std::less

Not always. By default, yes, but you can specify other comparators, e


std::map<int, int, std::greater> reversedMap;

Or at runtime:

   std::map<int, int> myMap(myPredicate);

Not really. There's a third type argument to std::map which specifies
the map's predicate type, with a default of std::less<T>. This
constructor takes an argument with the same type as the template's
predicate argument, so you can't pass arbitrary predicate objects. This
constructor is only useful with a user-defined predicate type that can
be initialized with something other than its default constructor.

Well, basically that just does mean that it depends on a runtime value.
Otherwise, you wouldn't need those constructor arguments in the first place.

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