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From: lgalfaso@gmail.com (Lucas Galfaso)
Newsgroups: comp.std.c++
Subject: Specialization of a template cast.
Date: Mon, 25 Apr 2005 01:25:20 GMT
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Xref: g2news1.google.com comp.std.c++:4584

Hi,
  From the standard, it is not clear how you should (a) create a
specialization from a template cast and (b) use a specific
specialization on a cast, ie


(A)

#include <iostream>
using namespace std;

class Foo {
 int m_foo;
public:
 Foo () : m_foo(0) {}

  template <typename _Ty>
  operator _Ty() const { cout << "This is a specialization on the type
" << typeid(_Ty).name() << endl; return (_Ty) m_foo; }

// How you write a specialization of the template cast above?
// template <> operator<int> int() const {return m_foo;} ?
// template <> operator int<int>() const {return m_foo;}?
// template <> operator int()<int> const {return m_foo;}?
};




(B)
#include <iostream>
using namespace std;

class Foo {
 int m_foo;
public:
 Foo () : m_foo(0) {}

  template <typename _Ty>
  operator _Ty() const { cout << "This is a specialization on the type
" << typeid(_Ty).name() << endl; return (_Ty) m_foo; }
  operator int() const { cout << "This is a standard cast to int" <<
endl; return m_foo; }
};



int main() {
 Foo foo;
  cout << (int)foo; // this will call the non-template version of the
cast, how you call the template version for the type int?

 return 0;
}


Thanks,
  Lucas Galfaso

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