How to call a templated function if it exists, and something else otherwise?

Question

I want to do something like

template 
void foo(const T& t) {
   IF bar(t) would compile
      bar(t);
   ELSE
      baz(t);
}

Answer 1

I think litb's solution works, but is overly complex. The reason is that he's introducing a function fallback::bar(...) which acts as a "function of last resort", and then goes to great lengths NOT to call it. Why? It seems we have a perfect behavior for it:

namespace fallback {
    template
    inline void bar(T const& t, ...)
    {
        baz(t);
    }
}
template
void foo(T const& t)
{
    using namespace fallback;
    bar(t);
}

But as I indicated in a comment to litb's original post, there are many reasons why bar(t) could fail to compile, and I'm not certain this solution handles the same cases. It certainly will fail on a private bar::bar(T t)