Re: [Boost-users] [MPL] implementing a "trait"?

7 Feb 2014

      -----Original Message-----
From: Boost-users [mailto:boost-users-bounces@lists.boost.org] On Behalf Of John M. Dlugosz

I want to mark certain types as being of a category of my invention (e.g. is_pretty), for purposes of using enable_if in the same manner is standard type traits such as is_arithmetic etc.

The trait will be false by default, and I'd declare something to nominate types that should be seen to have that trait.  I want it to follow inheritance; if B has been declared to be in my category, and C is derived from B, then C will also be in that category without needing to do anything more (though some way to turn it _off_ would be available).

-----End Original Message-----

The following works for me, but I am not sure if it is because of a working or a failing compiler:

Best wishes, Alex

#include <iostream>
#include <string>
#include 
#include <utility>

class A{}; //is_pretty;
class B{}; // don't know
class C : public A{}; // is_pretty because it derives from A
class D : public A{}; // not pretty even though it derives from A

// default because conversion pointer-to-bool has lowest overload precedence
std::false_type is_pretty_function(bool); 

std::true_type  is_pretty_function(A*);
std::false_type is_pretty_function(D*);

struct is_pretty_object 
{
  template <typename T>
  auto operator()(T&& t) -> decltype(is_pretty_function(std::forward<T>(t))) {
    return is_pretty_function(std::forward<T>(t));
  }
};

template<class T>
struct trait
{
  typedef typename std::result_of::type is_pretty;
};

int main()
{
  std::cout << "A looks " << (trait<A>::is_pretty::value ? "pretty": "nice" )<