Re: Q: STL priority_queue with boost shared_ptr

From:
Alan Johnson <alanwj@no.spam.stanford.edu>
Newsgroups:
comp.lang.c++
Date:
Mon, 05 Jun 2006 23:26:09 -0700
Message-ID:
<e63761$bn3$2@news.Stanford.EDU>
Alan Johnson wrote:

zl2k wrote:

hi, all
Here is what I want to do: to wrap my self defined class in a
shared_ptr and then insert it into the priority_queue. The
priority_queue should pump the least element of the self defined class
first. To make it simple, here is the toy code.

        bool operator<(const myClass & right) const { //the comparison is
"reversed" in order to pop smallest first
            return a > right.a;
        };


Okay, look's fine.

    priority_queue<shared_ptr<myClass> > pq;


Oops. You are using shared_ptr's operator<, not the one in myClass.

I was expecting the priority_queue to pop 1 first and 3 last. However,
it pops 3 first and 1 last.


You need to define a function object that does the comparison you want
when given two shared_ptr's. Example:

struct ptr_less
{
        bool operator()(const shared_ptr<myClass> & lhs,
                const shared_ptr<myClass> & rhs)
        {
                return *lhs < *rhs ;
        }
} ;

Or, if you want to be a bit more reusable, you could generalize this to
any type that can be dereferenced into a type that defines operator<.

Defining your priority_queue gets considerably more fun now. The second
template parameter is the container type. By default this is vector<T>,
so unless you have a reason just stick with that:

priority_queue<shared_ptr<myClass>,vector< shared_ptr<myClass> >,
ptr_less< shared_ptr<myClass> > > pq;

You might consider some typedefs to make that more readable.


Er. I left out the templated version of ptr_less, but the last line
there assumes that you are using it, so here it is:

template <typename Pointer>
struct ptr_less
{
         bool operator()(const Pointer & lhs,
                 const Pointer & rhs)
         {
                 return *lhs < *rhs ;
         }
} ;

--
Alan Johnson

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