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

Pete Becker <>
Fri, 21 Nov 2008 10:24:15 -0500
On 2008-11-21 06:34:27 -0500, Rolf Magnus <> said:

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.

It means that std::map<int, int> myMap(myPredicate) is an error unless
the type of myPredicate is std::less<int>, in which cast it's
irrelevant. In particular, it is not a runtime replacement for
std::map<int, int, std::greater<int>>.

Roundhouse Consulting, Ltd. ( Author of "The
Standard C++ Library Extensions: a Tutorial and Reference

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