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

joseph cook wrote:
A map is always sorted using std::less
Not always. By default, yes, but you can specify other comparators, eg:

std::map<int, int, std::greaterrev ersedMap;
Nov 20 '08 #1
6 5429
On Nov 20, 7:48*am, Juha Nieminen <nos...@thanks. invalidwrote:
joseph cook wrote:
A map is always sorted using std::less

* Not always. By default, yes, but you can specify other comparators, eg:

std::map<int, int, std::greaterrev ersedMap;
Or at runtime:

std::map<int, intmyMap(myPred icate);

Ali
Nov 20 '08 #2
On 2008-11-20 14:40:19 -0500, ac******@gmail. com said:
On Nov 20, 7:48Â*am, Juha Nieminen <nos...@thanks. invalidwrote:
>joseph cook wrote:
>>A map is always sorted using std::less

Â* Not always. By default, yes, but you can specify other comparators, e
g:
>>
std::map<int , int, std::greaterrev ersedMap;

Or at runtime:

std::map<int, intmyMap(myPred icate);
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)

Nov 20 '08 #3
Pete Becker wrote:
On 2008-11-20 14:40:19 -0500, ac******@gmail. com said:
>On Nov 20, 7:48 am, Juha Nieminen <nos...@thanks. invalidwrote:
>>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<in t, int, std::greaterrev ersedMap;

Or at runtime:

std::map<int, intmyMap(myPred icate);

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.
Nov 21 '08 #4
ac******@gmail. com wrote:
std::map<int, intmyMap(myPred icate);
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, MyPredicateType myMap(myPredica te);

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

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

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(myPred icate)myMap(myP redicate);
Nov 21 '08 #5
On 2008-11-21 06:34:27 -0500, Rolf Magnus <ra******@t-online.desaid:
Pete Becker wrote:
>On 2008-11-20 14:40:19 -0500, ac******@gmail. com said:
>>On Nov 20, 7:48 am, Juha Nieminen <nos...@thanks. invalidwrote:
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::greaterrev ersedMap;

Or at runtime:

std::map<in t, intmyMap(myPred icate);

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, intmyMap(myPred icate) 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<in t>>.

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

Nov 21 '08 #6
On 2008-11-21 08:46:16 -0500, Juha Nieminen <no****@thanks. invalidsaid:
ac******@gmail. com wrote:
>std::map<int , intmyMap(myPred icate);

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, MyPredicateType myMap(myPredica te);

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

std::map<int, int, bool(*)(int, int)myMap(myPre dicate);
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(myPre dicate);
>
This becomes even more awkward if the key and data types of the map
are something more complicated than int.
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)

Nov 21 '08 #7

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