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Subject: Re: [Boost-users] [MPL] implementing a "trait"?
From: Alex (alexhighviz_at_[hidden])
Date: 2014-02-08 12:44:50

-----Original Message-----
From: Boost-users [mailto:boost-users-bounces_at_[hidden]] On Behalf Of
Sent: 07 February 2014 18:48
To: boost-users_at_[hidden]
Subject: Re: [Boost-users] [MPL] implementing a "trait"?

-----Original Message-----
From: Boost-users [mailto:boost-users-bounces_at_[hidden]] 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-----

Now have a simpler solution:

#include <iostream>
#include <type_traits>
#include <utility>

class A{}; //pretty;
class B{}; // not pretty
class C : public A{}; // 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);

// set the traits for A and D
std::true_type is_pretty_function(A*);
std::false_type is_pretty_function(D*);

// use this struct to "read" the traits from the functions
template<class T> struct trait
  typedef decltype(is_pretty_function(std::declval<T*>())) is_pretty;

int main()
  std::cout << "A looks " << (trait<A>::is_pretty::value ? "pretty": "nice"
  std::cout << "B looks " << (trait<B>::is_pretty::value ? "pretty": "nice"
  std::cout << "C looks " << (trait<C>::is_pretty::value ? "pretty": "nice"
  std::cout << "D looks " << (trait<D>::is_pretty::value ? "pretty": "nice"
  return 0;

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