C++ std::priority_queue comparing using transformation

Question

Given class C, and function int f(C c), what is the proper idiom to create a std::priority_queue<C> which uses the standard < comparitor, not over C but over f(C)?

Write a predicate for use with the priority queue template, i.e. do *not* use operator `<` or the equivalent `std::less` in the queue. For implementing that predicate, you can of course use those. Note that this means you can *not* use `priority_queue` but that you will have to specify the third template parameter (and thus also the second), too. — Ulrich Eckhardt, Apr 12 '15 at 09:10
@UlrichEckhardt Does that require making a functor object? How else can I put the predicate into the template param? — SRobertJames, Apr 12 '15 at 09:36
Not an object but a type. You can also provide a function pointer, which is probably the easiest way to do it. You should be able to find examples for that if you look for examples involving the class templates `set` and `map` or the algorithms `find_if` and `sort`. — Ulrich Eckhardt, Apr 12 '15 at 10:10

score 0 · Answer 1 · edited May 23 '17 at 10:24

Refer 1

  ``// binary predicate function
    bool Compare1(C item1, C item2){
          return f(item1) > f(item2);
    }`   

    //`binary predicate functor
    class Compare2
    {
    public:
        bool operator() (C item1, C item2)
        {
            return f(item1) > f(item2);
        }
    };


int main()
{
    std::priority_queue<C, std::vector<C>, std::function<bool(C, C)>> pq1(Compare1);

    std::priority_queue<C, std::vector<C>, Compare2> pq2;
    return 0;
}

C++ std::priority_queue comparing using transformation

1 Answers1