Hi!
Why do you need an enable_if if you exactly know the type T? In that case you can use the temmplate specialization.
template<>
void f(const std::string&, bool) {...}
template<>
void f(const std::string&, foo) {...}
template<typename T>
void f(const std::string&, typename enable_if<boost::is_enum<T>, T>::type t) {...}
Suppose I have a template function
template<typename T> void f(const std::string&, T);
which I want there to be a couple of different variants of.
Version 1 - Enabled if and only if boost::is_same<T, bool>
Version 2 - Enabled if and only if boost::is_same<T, foo> (foo is a class)
Version 3 - Enabled if and only if boost::is_enum<T>
Version 4 - Otherwise
Is there an easy way to express this? It seems I run into a bunch of overload ambiguity problems. I'm sure I can get around this by adding to every version the negation of all the other conditions, but this seems unnecessarily complicated.
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