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

joseph cook wrote:Not always. By default, yes, but you can specify other comparators, eg:

std::map<int, int, std::greaterreversedMap;

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

On Nov 20, 7:48*am, Juha Nieminen <nos...@thanks.invalidwrote:Or at runtime:

std::map<int, intmyMap(myPredicate);

Ali

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

On 2008-11-20 14:40:19 -0500, acehreli@gmail.com said:
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.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

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

Pete Becker wrote:
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.

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

acehreli@gmail.com wrote:I don't think you can do that because the comparator template
parameter is set by default to std::less, and unless myPredicate casts
implicitly to type std::less, that won't work. You have to do it like:

std::map<int, int, MyPredicateTypemyMap(myPredicate);

If 'myPredicate' is a function, the syntax becomes awkward:

std::map<int, int, bool(*)(int, int)myMap(myPredicate);

This becomes even more awkward if the key and data types of the map
are something more complicated than int.

The next standard will offer a tool to alleviate the problem:

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

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

On 2008-11-21 06:34:27 -0500, Rolf Magnus <ramagnus@t-online.desaid:
It means that std::map<int, intmyMap(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>>.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

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

On 2008-11-21 08:46:16 -0500, Juha Nieminen <nospam@thanks.invalidsaid:
Well, yes, and as we've seen in many Ginsu knife commercials, a normal
knife can't slice a tomato. The way to write this code is, of course,
with appropriate typedefs:

typedef bool (*pred)(int,int);
std::map<int, int, predmyMap(myPredicate);
Not at all. Again, typedefs.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)