Variadic template: Perfect forwarding of integer parameter to lambda

Question

There are similar questions, but I did not find an answer that works for my problem.

Consider the following code:

#include <cassert>
#include <functional>
#include <iostream>
#include <memory>
#include <utility>

class TestClass
{
public:
   TestClass( int value): mValue( value) { }
private:
   int  mValue;
};

template< typename T> class DeferredCreator
{
public:
   template< class... Args> DeferredCreator( Args&&... args):
      mpCreator( [=]() -> T*
         { return new T( std::forward< Args>( args)...);  }
      ),
      mpObject()
   { }

   T* get() {
      if (mpObject == nullptr)
         mpObject.reset( mpCreator());
      return mpObject.get();
   }
private:
   std::function< T*( void)>  mpCreator;
   std::unique_ptr< T>        mpObject;
};


int main() {
   DeferredCreator< int>  dcInt( 42);

   assert( dcInt.get() != nullptr);
   return 0;
}

The idea is that the class DeferredCreator creates an object only when it is really needed. I got this work e.g. for strings, but I can't figure out how to pass a simple integer into my lambda.

The error message I get is:

prog.cpp:19:26: error: no matching function for call to 'forward'
         { return new T( std::forward< Args>( args)...);  }
                         ^~~~~~~~~~~~~~~~~~~
prog.cpp:36:27: note: in instantiation of function template specialization 'DeferredCreator<int>::DeferredCreator<int>' requested here
   DeferredCreator< int>  dcInt( 42);
                          ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/6.3.0/../../../../include/c++/6.3.0/bits/move.h:76:5: note: candidate function not viable: 1st argument ('const int') would lose const qualifier
    forward(typename std::remove_reference<_Tp>::type& __t) noexcept
    ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/6.3.0/../../../../include/c++/6.3.0/bits/move.h:87:5: note: candidate function not viable: 1st argument ('const int') would lose const qualifier
    forward(typename std::remove_reference<_Tp>::type&& __t) noexcept
    ^
2 errors generated.

I already tried to use decltype( args) as template argument for std::forward<>, but that did not help.

The code is also available here: https://ideone.com/MIhMkt

Answer 1

args... is constant because a lambda's call operator is implicitly const. So, if you make your lambda mutable, then it works:

[=]() mutable -> T*
     { return new T( std::forward< Args>( args)...);  }

The reason it didn't work with decltype(args) is that the types themselves are not const, just the call operator.

Answer 2

The operator() of the closure type generated by your lambda expression is const-qualified. std::forward can attempt to move args..., which are data members of the closure. const objects cannot be moved.

You can mark your lambda as mutable:

  mpCreator( [=]() mutable -> T*
     { return new T( std::forward< Args>( args)...);  }
  ),

This removes the implicit const qualfiier from the closure type's generated operator().

live example on wandbox.org