Define priority_queue with lambda in header file












1















I have a question about defining a priority_queue pointer in header file and initializing it in source file.



header file



std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, ?> *readyQueue;


source file



auto readyQueueComparator = (const std::shared_ptr<Process> &first, const std::shared_ptr<Process> &second) {
return true;
};
readyQueue = new std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, decltype(readyQueueComparator)>;


Basically, I want to implement a structure like this but I couldn't figure out how should I define priority queue in header file in this format.



Thanks.










share|improve this question

























  • I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.

    – NathanOliver
    Nov 19 '18 at 21:24











  • That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.

    – SergeyA
    Nov 19 '18 at 21:28













  • @SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?

    – Mert Akozcan
    Nov 19 '18 at 21:58











  • @MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.

    – SergeyA
    Nov 19 '18 at 22:00


















1















I have a question about defining a priority_queue pointer in header file and initializing it in source file.



header file



std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, ?> *readyQueue;


source file



auto readyQueueComparator = (const std::shared_ptr<Process> &first, const std::shared_ptr<Process> &second) {
return true;
};
readyQueue = new std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, decltype(readyQueueComparator)>;


Basically, I want to implement a structure like this but I couldn't figure out how should I define priority queue in header file in this format.



Thanks.










share|improve this question

























  • I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.

    – NathanOliver
    Nov 19 '18 at 21:24











  • That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.

    – SergeyA
    Nov 19 '18 at 21:28













  • @SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?

    – Mert Akozcan
    Nov 19 '18 at 21:58











  • @MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.

    – SergeyA
    Nov 19 '18 at 22:00
















1












1








1








I have a question about defining a priority_queue pointer in header file and initializing it in source file.



header file



std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, ?> *readyQueue;


source file



auto readyQueueComparator = (const std::shared_ptr<Process> &first, const std::shared_ptr<Process> &second) {
return true;
};
readyQueue = new std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, decltype(readyQueueComparator)>;


Basically, I want to implement a structure like this but I couldn't figure out how should I define priority queue in header file in this format.



Thanks.










share|improve this question
















I have a question about defining a priority_queue pointer in header file and initializing it in source file.



header file



std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, ?> *readyQueue;


source file



auto readyQueueComparator = (const std::shared_ptr<Process> &first, const std::shared_ptr<Process> &second) {
return true;
};
readyQueue = new std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, decltype(readyQueueComparator)>;


Basically, I want to implement a structure like this but I couldn't figure out how should I define priority queue in header file in this format.



Thanks.







c++






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 19 '18 at 21:23









NathanOliver

90.8k15126190




90.8k15126190










asked Nov 19 '18 at 21:21









Mert AkozcanMert Akozcan

6317




6317













  • I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.

    – NathanOliver
    Nov 19 '18 at 21:24











  • That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.

    – SergeyA
    Nov 19 '18 at 21:28













  • @SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?

    – Mert Akozcan
    Nov 19 '18 at 21:58











  • @MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.

    – SergeyA
    Nov 19 '18 at 22:00





















  • I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.

    – NathanOliver
    Nov 19 '18 at 21:24











  • That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.

    – SergeyA
    Nov 19 '18 at 21:28













  • @SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?

    – Mert Akozcan
    Nov 19 '18 at 21:58











  • @MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.

    – SergeyA
    Nov 19 '18 at 22:00



















I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.

– NathanOliver
Nov 19 '18 at 21:24





I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.

– NathanOliver
Nov 19 '18 at 21:24













That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.

– SergeyA
Nov 19 '18 at 21:28







That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.

– SergeyA
Nov 19 '18 at 21:28















@SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?

– Mert Akozcan
Nov 19 '18 at 21:58





@SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?

– Mert Akozcan
Nov 19 '18 at 21:58













@MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.

– SergeyA
Nov 19 '18 at 22:00







@MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.

– SergeyA
Nov 19 '18 at 22:00














1 Answer
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The short answer is don't use a lambda for this. Define a custom callable type instead. For example:



struct OrderByProcessPriority
{
bool operator()(
std::shared_ptr<Process> const& first,
std::shared_ptr<Process> const& second) const;
}

using ProcessQueue = std::priority_queue<
std::shared_ptr<Process>,
std::vector<std::shared_ptr<Process>>,
OrderByProcessPriority>;

ProcessQueue *readyQueue;


Then, in your cpp file, create the instance like this:



readyQueue = new ProcessQueue{};





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    1 Answer
    1






    active

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    active

    oldest

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    active

    oldest

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    3














    The short answer is don't use a lambda for this. Define a custom callable type instead. For example:



    struct OrderByProcessPriority
    {
    bool operator()(
    std::shared_ptr<Process> const& first,
    std::shared_ptr<Process> const& second) const;
    }

    using ProcessQueue = std::priority_queue<
    std::shared_ptr<Process>,
    std::vector<std::shared_ptr<Process>>,
    OrderByProcessPriority>;

    ProcessQueue *readyQueue;


    Then, in your cpp file, create the instance like this:



    readyQueue = new ProcessQueue{};





    share|improve this answer




























      3














      The short answer is don't use a lambda for this. Define a custom callable type instead. For example:



      struct OrderByProcessPriority
      {
      bool operator()(
      std::shared_ptr<Process> const& first,
      std::shared_ptr<Process> const& second) const;
      }

      using ProcessQueue = std::priority_queue<
      std::shared_ptr<Process>,
      std::vector<std::shared_ptr<Process>>,
      OrderByProcessPriority>;

      ProcessQueue *readyQueue;


      Then, in your cpp file, create the instance like this:



      readyQueue = new ProcessQueue{};





      share|improve this answer


























        3












        3








        3







        The short answer is don't use a lambda for this. Define a custom callable type instead. For example:



        struct OrderByProcessPriority
        {
        bool operator()(
        std::shared_ptr<Process> const& first,
        std::shared_ptr<Process> const& second) const;
        }

        using ProcessQueue = std::priority_queue<
        std::shared_ptr<Process>,
        std::vector<std::shared_ptr<Process>>,
        OrderByProcessPriority>;

        ProcessQueue *readyQueue;


        Then, in your cpp file, create the instance like this:



        readyQueue = new ProcessQueue{};





        share|improve this answer













        The short answer is don't use a lambda for this. Define a custom callable type instead. For example:



        struct OrderByProcessPriority
        {
        bool operator()(
        std::shared_ptr<Process> const& first,
        std::shared_ptr<Process> const& second) const;
        }

        using ProcessQueue = std::priority_queue<
        std::shared_ptr<Process>,
        std::vector<std::shared_ptr<Process>>,
        OrderByProcessPriority>;

        ProcessQueue *readyQueue;


        Then, in your cpp file, create the instance like this:



        readyQueue = new ProcessQueue{};






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 19 '18 at 21:31









        Peter RudermanPeter Ruderman

        10.2k2352




        10.2k2352






























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