Is it legal to declare a constexpr initializer_list object?

dyp picture dyp · Apr 17, 2013 · Viewed 8.5k times · Source

As a question that came up during the discussion of this SO question:

Is it legal, maybe with N3471, to declare a constexpr std::initializer_list object? Example:

constexpr std::initializer_list<int> my_list{};

Why I think it may not be legal: initializer_list would have to be a literal type; but are there any guarantees that it is a literal type?

Citations from N3485.

[dcl.constexpr]/9:

A constexpr specifier used in an object declaration declares the object as const. Such an object shall have literal type and shall be initialized.

literal types requirements, [basic.types]/10, sub-bullet class types:

  • a class type (Clause 9) that has all of the following properties:
    • it has a trivial destructor,
    • every constructor call and full-expression in the brace-or-equal-initializers for non-static data members (if any) is a constant expression (5.19),
    • it is an aggregate type (8.5.1) or has at least one constexpr constructor or constructor template 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.

Bonus points ;) for answering if

constexpr std::initializer_list<int> my_list = {1,2,3,4,5};

is legal (with references). Though I think this is covered by the above + [dcl.init.list]/5

Answer

dyp picture dyp · Apr 17, 2013

Update: The situation got a bit more complicated after the resolution of CWG DR 1684 removed the requirement quoted below. Some more information can be found in this discussion on the std-discussion mailing list and in the related question Why isn't `std::initializer_list` defined as a literal type?


[decl.constexpr]/8:

A constexpr specifier for a non-static member function that is not a constructor declares that member function to be const (9.3.1). [...] The class of which that function is a member shall be a literal type (3.9).

Therefore, the changes of N3471 guarantee std::initializer_list will be a literal type.


Note the constexpr ctor alone doesn't require std::initializer_list to be a literal type, see [dcl.constexpr]/4+8. Side note: An object of non-literal type with constexpr ctor can be initialized during constant initialization [basic.start.init]/2] (part of static initialization, performed before any dynamic initialization).