I suspect boost::optional's get_value_or
was deprecated because it is unsafe if an rvalue is passed as the default
parameter. However, it is occasionally useful to be able to reference the optional value or a default alternative.
Is the following safe?
template<typename T>
T const& get_reference_or(boost::optional<T> const& opt, T const& alt)
{
if (opt) return opt.get();
else return alt;
}
template<typename T>
T const& get_reference_or(boost::optional<T> const&, T&&) = delete;
As written, your code has problems with type deduction, because T
is deducible from both arguments. But suppose you actually made T
only deducible from the optional:
template <class T>
struct NonDeducedHelper { using type = T; };
template <class T>
using NonDeduced = typename NonDeducedHelper<T>::type;
template<typename T>
T const& get_reference_or(boost::optional<T> const& opt, NonDeduced<T> const& alt)
{
if (opt) return opt.get();
else return alt;
}
template<typename T>
T const& get_reference_or(boost::optional<T> const&, NonDeduced<T>&&) = delete;
Then, the code is almost safe, since when a non-const
rvalue is used as the default for get_reference_or
, it will attempt to use the deleted overload and fail to compile. However, to be 100% safe, you should also delete the overload for const
rvalues:
template<typename T>
T const& get_reference_or(boost::optional<T> const&, NonDeduced<T> const&&) = delete;