Here is described the nullopt_t
and nullopt
for the optional
object proposed for c++:
struct nullopt_t{see below}; constexpr nullopt_t nullopt(unspecified);
[...] Type nullopt_t shall not have a default constructor. It shall be a literal type. Constant nullopt shall be initialized with an argument of literal type.
The reason for this is explained in the The op = {} syntax chapter of the document: for the op = {}
to be unambiguous some tricks have to be adopted, one of which is that nullopt_t
must not be default constructible.
My question is about what does the literal type means here? I found this SO post. So it looks to me that just another empty class would do. Could it also be a constructor taking a int
?
What would be a minimal conforming nullopt_t
class look like?
Something like this:
struct nullopt_t_construct_tag_t{};
struct nullopt_t {
nullopt_t() = delete; // I know declaring it as deleted is redundant
constexpr nullopt_t(nullopt_t_construct_tag_t) {};
};
constexpr nullopt_t nullopt(nullopt_t_construct_tag_t{});
Or this:
struct nullopt_t {
nullopt_t() = delete;
constexpr nullopt_t(int) {};
};
constexpr nullopt_t nullopt(0);
A minimal implementation is
struct nullopt_t {
constexpr nullopt_t(int) {}
};
No default constructor will be implicitly declared, [class.ctor]/4:
If there is no user-declared constructor for class
X
, a constructor having no parameters is implicitly declared as defaulted (8.4).
... and nullopt_t
can be constructed from int
, a literal type.
Note that in your code a default constructor exists, though being defined as deleted.
The above definition does meet the requirements for a literal type:
A type is a literal type if it is:
(10.5) — a class type (Clause 9) that has all of the following properties:
- it has a trivial destructor,
- it is an aggregate type (8.5.1) or has at least one
constexpr
constructor [..] that is not a copy or move constructor, and- all of its non-static data members and base classes are of non-volatile literal types.
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