Re: template matching
"assaf" <assaflavie@gmail.com> writes:
What should bar() output and why?
class Base {};
class Derived : public Base {};
template <typename T> void foo(const T&) { cout << "a"; }
template <> void foo(const Base&) { cout << "b"; }
void bar()
{
foo(Derived());
}
"a". The set of candidates consists of the explicit specialization for
Base and the specialization for Derived generated from the base
template. The latter is the better match.
"a" is always printed. I want a specialization of foo that would
handle all descendants of Base, while all other types are handled by
the primary template function. Is that possible?
Not directly, I'm afraid.
Something like
template <bool>
void print()
{
std::cout << "a";
}
template <>
void print<true>()
{
std::cout << "b";
}
template <typename T>
void foo(const T &)
{
Print<boost::is_base_of<Base,T>::value>::execute();
}
should work, though.
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