As T.C. demonstrated with some links in a comment, the standard is not quite clear on this; a similar problem arises with trailing return types using decltype(memberfunction()). The central problem is that class members are generally not considered to be declared until after the class in which they’re declared is complete. Thus, regardless of the … Read more
For the record, the static constexpr version will work like you’d expected in C++17. From N4618 Annex D.1 [depr.static_constexpr]: D.1 Redeclaration of static constexpr data members [depr.static_constexpr] For compatibility with prior C++ International Standards, a constexpr static data member may be redundantly redeclared outside the class with no initializer. This usage is deprecated. [Example: struct … Read more
From memory, member function bodies are evaluated only once the class has been completely defined. static constexpr int bah = static_n_items(); forms part of the class definition, but it’s referring to a (static) member function, which cannot yet be defined. Solution: defer constant expressions to a base class and derive from it. e.g.: struct Item_id_base … Read more
It is possible to detect if a given function-call expression is a constant expression, and thereby select between two different implementations. Requires C++14 for the generic lambda used below. (This answer grew out this answer from @Yakk to a question I asked last year). I’m not sure how far I’m pushing the Standard. This is … Read more
clang is correct, note the HEAD revision of gcc accepts also accepts this code. This is a well-formed constexpr function, as long as there is value for the argument(s) that allows the function to be evaluated as a core constant expression. In your case 1 is such a value. This is covered in the draft … Read more
Clang seems to be in the right. When accessing a static member with the member access syntax [class.static/1]: A static member s of class X may be referred to using the qualified-id expression X::s; it is not necessary to use the class member access syntax to refer to a static member. A static member may … Read more
You tell the compiler, that addFunc would be a constexpr. But it depents on parameters, that are not constexpr itself, so the compiler already chokes on that. Marking them const only means you are not going to modify them in the function body, and the specific calls you make to the function are not considered … Read more
Microsoft publishes a C++11 compatibility table, under which constexpr is clearly marked as not being available in Visual Studio 2013. The November 2013 CTP has it, though. Source: Google visual studio constexpr
constexpr variable is guaranteed to have a value available at compile time. whereas static const members or const variable could either mean a compile time value or a runtime value. Typing constexpr express your intent of a compile time value in a much more explicit way than const. One more thing, in C++17, constexpr static … Read more
This is because y.n is not odr-used and therefore does not require an access to y.n the rules for odr-use are covered in 3.2 and says: A variable x whose name appears as a potentially-evaluated expression ex is odr-used unless applying the lvalue-to-rvalue conversion (4.1) to x yields a constant expression (5.19) that does not … Read more