Re: Templated Casting operators

From:
Narinder <narinder.claire@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Sat, 9 Jul 2011 07:18:58 -0700 (PDT)
Message-ID:
<b781d567-f97a-4951-b3fb-b6702884a8d8@h12g2000vbx.googlegroups.com>
On Jul 9, 12:34 pm, Kai-Uwe Bux <jkherci...@gmx.net> wrote:

Narinder wrote:

On Jul 8, 11:06 pm, Kai-Uwe Bux <jkherci...@gmx.net> wrote:
At the moment, I don't see why.


That is indeed curious.
However on my issue, I think I am expecting something from C++ .. when
I obviously should know better, "C++ (101): C++ doesn't dispatch on
the return type".


That, I think, is not the entire picture. There is clause [14.8.2.3] that
deals with deducing template arguments for templated conversion operators=

..

Here, the result type desired by the conversion is taken into account.
Howerver, reading that clause, I find it hard to tell whether the deducti=

on

should fail because of ambiguity, whether the const version should be
instantiated, or whether the non-const version should be chosen.

My question stems from wanting to pass parameters to a function via an
intermediary which holds references to the values to be passed. And
this would be done at runtime.

Given :

void f( double x);

I want to be able to do:

const double x=3.142;
intermediary v = x; // so v will hold a
std::tr1::reference_wrapper<const double>
f(v); // I suspect gcc will complain becaus=

e v is unable

to give up its const ref
                    // but as we have seen gcc call=

s operator

double&()


Hm, the following compiles:

#include <iostream>

void foo ( double arg ) {
  std::cout << arg << "\n";

}

  template < typename T >
  class reference_wrapper {

    T* the_ptr;

  public:

    typedef T type;
    typedef T& reference;

    reference_wrapper ( reference t_ref )
      : the_ptr ( &t_ref )
    {}

    operator reference ( void ) {
      return ( *the_ptr );
    }

    reference get ( void ) {
      return ( *the_ptr );
    }

 };

int main ( void ) {
  double const pi = 3.14159;
  reference_wrapper< double const > x = pi;
  foo( x );

}

My way around this will be to have the user provide a precise boost
typelist embodying the function signature:

boost::mpl::vector<double>

and use this to dispatch correctly.


Huh?

Best,

Kai-Uwe Bux


Consider the following functions:

f_ref(double&) // may be a member function that keeps the
rw ref
f_const_ref(const double&) // maybe a memeber function that keeps the
ro ref
f_val(double) // expects the value byval

you have one 'variant'

variant v = 3.142;

double x=3.142;
variant v2 = x;

const double xx=3.142
variant v3 = 3.142

I want to deliver (since it doesn't have to be done by casting) v2 to
f_ref, but not v & v3
I want to deliver v2 & v3 to f_const_ref but not v
and i want to be able to deliver all of them to f_val

Moreover my variants will be initialised at runtime.

I have sketched (but compiles and executes ) what I think is a
solution.
(I have used boost::mpl::vector to be able to generalise to more than
one parameter)

-------------------------------
#include<iostream>
#include<boost/ref.hpp>
#include<boost/variant.hpp>
#include<boost/mpl/vector.hpp>
#include<boost/mpl/at.hpp>

using namespace std;

double pi = 3.142;
typedef boost::reference_wrapper<double> ref;
typedef boost::reference_wrapper<const double> const_ref;

struct klass
{
    klass(double & x):innerVariant(boost::ref(x)){}
    klass(const double & x):innerVariant(boost::cref(x)){}

    template<class T>
    operator T& ()
    {
        cout << " ... casting to double& ";
        return boost::get<ref>(innerVariant);
    }

    template<class T>
    operator const T&()
    {
        cout << " ... casting to const double& ";
        if(boost::get<ref>(&innerVariant))
        {
            return boost::get<ref>(innerVariant);
        }
        return boost::get<const_ref>(innerVariant);

    }

    const double &get_by_val()
    {
        return (const double&)(*this);
    }

private:
    boost::variant<ref,const_ref > innerVariant;
};

template<class T,class U>
struct innerCast
{
    static const T & get(U &k)
    {
        return k.get_by_val();
    }
};

template<class T,class U>
struct innerCast<T&,U>
{
    static T & get(U &k)
    {
        return k;
    }
};

template<class T,class U>
struct innerCast<const T&,U>
{
    static const T & get(U &k)
    {
        return k;
    }
};

template<class T>
struct cast
{
    typedef typename boost::mpl::at_c<typename T::param_type,0>::type
ret_type;

    static ret_type get(klass &k)
    {
        return innerCast<ret_type,klass>::get(k);
    }

};

/////////////////////////////////////////
// Below is what would be the user code
/////////////////////////////////////////

struct f
{
    typedef boost::mpl::vector<double&>::type param_type;
    static void call(double &x)
    {
        cout << " double &x\n";
    }
};

struct fc
{
    typedef boost::mpl::vector<const double&> param_type;
    static void call(const double &x)
    {
        cout << " const double &x\n";
    }
};

struct ft
{
    typedef boost::mpl::vector<double> param_type;
    static void call(double x)
    {
        cout << " double\n";
    }

};

int main()
{

    try
    {

        double x(3.142);

        klass k_non_const(x);
        klass k_const(3.142);

        f::call( cast<f>::get(k_non_const) );
        fc::call( cast<fc>::get(k_non_const));
        ft::call( cast<ft>::get(k_non_const));

        fc::call( cast<fc>::get(k_const));
        ft::call( cast<ft>::get(k_const)); // this one *NO LONGER*
throws with gcc
    }
    catch(const std::exception &err)
    {
        cout << err.what() << "\n";
    }

}

---------------------------------------------

Maybe some of the class names have alot of scope to be more
meaningful :-)

Best Regards
N

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