Re: Sorting a map<T*>

From:

Pete Becker <pete@versatilecoding.com>

Newsgroups:

comp.lang.c++

Date:

Mon, 07 May 2007 21:00:59 -0400

Message-ID:

<L-OdnXFcTZJWVqLbnZ2dnUVZ_qHinZ2d@giganews.com>

Greg Herlihy wrote:

On 5/7/07 5:01 PM, in article pan.2007.05.08.00.01.10@yahoo.de, "Markus
Schoder" <a3vr6dsg-usenet@yahoo.de> wrote:

On Mon, 07 May 2007 19:45:00 -0400, Pete Becker wrote:

Markus Schoder wrote:

On Mon, 07 May 2007 16:11:52 -0700, Michael wrote:

On May 7, 3:01 pm, "barcaroller" <barcarol...@music.net> wrote:

I have a set<T*> that stores pointers to objects. How can I tell
set<T*> to use the objects' operator<() and not the value of the
pointers for sorting?

There's a way to create a set with two parameters: set<MyType,
key_comp>, where
key_comp is a key comparison function you define. For example,
set<int, less<int> >.

So then you'd just need to provide the appropriate comparison function
that uses
the objects' operator<(), suitably wrapped.

Actually you need a class with an appropriate operator(). A plain
function won't do here since you cannot pass it as a template
parameter.

Sure you can. Try it.

No you cannot. I guess you are confusing this with algorithms like
std::lower_bound where you can specify a predicate as a _function_
parameter. As a template parameter for std::set or std::map this does not
work. Think about it -- it needs to be a type.

But std::set's Compare type may be a function type - it does not need to be
a class type that has an overloaded operator() defined. So Pete's objection
is correct - and here is some code to demonstrate:

    #include <set>

    bool MyComp(int *a, int *b)
    {
        return *a < *b;
    }

    int main()
    {
       std::set<int*, bool (*)(int*, int*)> s(MyComp);

       s.insert( new int(1) );
       s.insert( new int(2) );
       ...
    }

You're right. Spoke too quickly.

--

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