peter55555
peter55555

Reputation: 1475

How to pass other template parameter when template class uses parameter pack?

I would like to create template class that implements print() method for each type passed as template parameters.

Something like that:

class Interface
{
public:
    virtual ~Interface() = default;
    virtual void print(int) = 0;
    virtual void print(double) = 0;
};
X x<int, double, Interface>;

class X has public method void print() and it works.

The whole code below:

#include <iostream>
#include <type_traits>

struct Printer
{
    void print(int i) {std::cout << i << std::endl; }
    void print(double d) {std::cout << d << std::endl; } 
};

class Interface
{
public:
    virtual ~Interface() = default;
    virtual void print(int) = 0;
    virtual void print(double) = 0;
};

template <typename... Args>
class X;

template <typename Interface>
class X<Interface> : public Interface
{
    static_assert(std::is_abstract<Interface>::value, "Last argument should be an interface");

public:
    X(Printer printer) {}
    using Interface::print;
};

template <typename Arg, typename... Args>
class X<Arg, Args...> : public X<Args...>
{
    using Parent = X<Args...>;

public:
    using Parent::print;

    X(Printer printer_): Parent(printer), printer{printer_} {}
    void print(Arg arg) override { printer.print(arg); }

private:
    Printer printer;
};

int main()
{
    Printer printer;
    X<double, int, Interface> x(printer);
    x.print(5);
}

As you see class X uses Printer class but the problem is that I would like to have Printer as a template parameter...

Is it possible? How to do that?

Upvotes: 1

Views: 426

Answers (2)

max66
max66

Reputation: 66200

As you see class X uses Printer class but the problem is that I would like to have Printer as a template parameter...

Is it possible? How to do that?

Sorry but... I don't see the problem (with a great simplification suggested by Story Teller: place a single Printer object in the ground case case)

template <typename...>
class X;

template <typename Printer, typename Interface>
class X<Printer, Interface> : public Interface
 {
   static_assert(std::is_abstract<Interface>::value,
                 "Last argument should be an interface");

   public:
      X (Printer p0) : printer{p0}
       { }

      using Interface::print;  // why?

   protected:
      Printer printer;
 };

template <typename Printer, typename Arg, typename... Args>
class X<Printer, Arg, Args...> : public X<Printer, Args...>
 {
   using Parent = X<Printer, Args...>;

   public:
      using Parent::print;
      using Parent::printer;

      X(Printer printer_): Parent{printer_} {}

      void print(Arg arg) override { printer.print(arg); }
 };

// ....

X<Printer, double, int, Interface> x(printer);

Off topic: attention: you're using printer uninitialized

X(Printer printer_): Parent(printer), printer{printer_} {}

I suppose you should write Parent(printer_)

Upvotes: 2

Richard Hodges
Richard Hodges

Reputation: 69882

  • Assuming that the polymorphic interface is required.
  • polymorphism reduces the value of variadic template expansion
  • preserving the deferral of action to an encapsulated printer

A possible solution:

#include <iostream>
#include <type_traits>

// Abstract interface
class PrintInterface
{
public:
    virtual ~PrintInterface() = default;
    virtual void print(int) = 0;
    virtual void print(double) = 0;
};

// An implmentation of PrintInterface that defers to PrinterType
template<class PrinterType>
class ImplementPrintInterface : public PrintInterface
{
public:
    ImplementPrintInterface(PrinterType printer)
    : printer_(std::move(printer))
    {}

    virtual void print(int x) override
    {
        printer_.print(x);
    }

    virtual void print(double x) override
    {
        printer_.print(x);
    }

private:

    PrinterType printer_;
};

// An implementation of a thing that prints ints and doubles.
// This happens to match PrintInterface but there is no inheritance
struct Printer
{
    void print(int i) {std::cout << i << std::endl; }
    void print(double d) {std::cout << d << std::endl; } 
};

// X *is a* PrinterInterface that *uses a* PrinterType
template <typename PrinterType>
class X : public ImplementPrintInterface<PrinterType>
{
public:
    X(PrinterType printer = PrinterType())
    : ImplementPrintInterface<PrinterType>(std::move(printer))
    {}

};

int main()
{
    Printer printer;
    X<Printer> x(printer);
    x.print(5);
}

Upvotes: 0

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