[Boost-users] puzzle with enable_if

Subject: [Boost-users] puzzle with enable_if
From: John Dlugosz (JDlugosz_at_[hidden])
Date: 2010-04-14 16:46:38

I'm using the Microsoft 2008 compiler.

The template

 template <typename T, typename D>
 struct is_proper_TD_types : std::tr1::integral_constant<bool,
    is_Universal_time_API<T>::value &&
      (std::tr1::is_same<typename T::duration_t,D>::value || 
       std::tr1::is_same<duration,D>::value || 
       std::tr1::is_same<linear_duration,D>::value || 
       is_dimensional_time<D>::value)
   > {};

gives an error "'T': must be a class or namespace when followed by '::'" when T is int.  This kind of thing is mentioned in the enable_if documentation as the motivation for the "lazy" forms, because the compiler doesn't apply SFINAE for other templates.

But, riddle me this:  I created that line as an attempt to refactor the following, which DOES work OK:

 template <typename T, typename D>
 typename boost::enable_if_c<
   is_Universal_time_API<T>::value && 
   (std::tr1::is_same<typename  T::duration_t,D>::value || 
    std::tr1::is_same<duration,D>::value || 
    std::tr1::is_same<linear_duration,D>::value || 
    is_dimensional_time<D>::value)
,T>::type
   operator+ (const T& t1, const D& d1)
 {
 // blah blah


Basically, I just copied the bulk of the first argument of enable_if_c in the second listing to make a named metaprogramming predicate that I could use in several places.

I don't see anything in your implementation of enable_if_c that would cause it to work:  the argument is a template bool constant in both cases.  Why does the bottom one work, when the top one doesn't?

--John








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