[Boost-users] Lambda return type deduction off of the generated lambda fctor (is it possible?)

From: Brian Braatz (brianb_at_[hidden])
Date: 2004-12-21 15:05:54

I wrote up the simple example that shows what I am trying to figure out

 

Basically, I am trying to pull the deduced return type (or explicitly
passed as a template param in lambda bind()) in my template:

 

Any help or direction is greatly appreciated

 

      template <typename CURRIEDTYPE>

      // Id class for currying type without creating the object

      class Id

      {

            public:

                  typedef CURRIEDTYPE TYPE;

      };

      // Explictly pass IN a return type

      template < typename TFUNCTOR, typename TRET>

      TRET CallFctorPassInRet ( TFUNCTOR & functor, Id<TRET> & )

      {

            Id<TRET>::TYPE TRet;

            TRet = functor();

            return TRet;

      }

 

      std::string ReturnAString()

      {

            return "this is a string";

      }

 

      // try to get return type from the lambda fctor

      // Below does not compile- but it is along the lines of what I 

      // WANT to do- note that I am trying to get from the Lambda
functor the deduced return type

      template < typename TFUNCTOR>

      TFUNCTOR::result_type CallFctorDeduceRet( TFUNCTOR & functor)

      {

            return functor();

      }

      

      

      void Test_LambdaRetDeduction()

      {

            std::cout << "CallFctorPassInRet returns [" <<
CallFctorPassInRet( bind(&ReturnAString), Id<std::string>()) << "]" <<
std::endl;

            

            // below is what I want to make work

             std::cout << "CallFctorDeduceRet returns [" <<
CallFctorDeduceRet( bind<std::string>(&ReturnAString)) << "]" <<
std::endl;

            }