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

joseph cook wrote:

Quote:

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::greaterreversedMap;

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

Quote:

joseph cook wrote:

Quote:

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::greaterreversedMap;

Or at runtime:

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

Ali

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

Quote:

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

Quote:

>joseph cook wrote:

Quote:

>>A map is always sorted using std::less

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

g:

Quote:

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

>
Or at runtime:
>
std::map<int, intmyMap(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.

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

Pete Becker wrote:

Quote:

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

Quote:

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

Quote:

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

>g:

Quote:

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

>>
>Or at runtime:
>>
> std::map<int, intmyMap(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.

acehreli@gmail.com wrote:

Quote:

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

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);

On 2008-11-21 06:34:27 -0500, Rolf Magnus <ramagnus@t-online.desaid:

Quote:

Pete Becker wrote:
>

Quote:

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

Quote:

>>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::greaterreversedMap;
>>>
>>Or at runtime:
>>>
>>std::map<int, intmyMap(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, 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)

On 2008-11-21 08:46:16 -0500, Juha Nieminen <nospam@thanks.invalidsaid:

Quote:

acehreli@gmail.com wrote:

Quote:

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

>
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);

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);

Quote:

>
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)

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

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

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

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

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

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

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

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