I have the following code:

static constexpr const char*const myString = "myString";

Could you please explain what is the difference from: 

static const char*const myString = "myString";

What's new we have with constexpr in this case?

2

1 Answer

The difference is described in the following quote from the C++ Standard (9.4.2 Static data members)

3 If a non-volatile const static data member is of integral or enumeration type, its declaration in the class definition can specify a brace-or-equal-initializer in which every initializer-clause that is an assignmentexpression is a constant expression (5.19). A static data member of literal type can be declared in the class definition with the constexpr specifier; if so, its declaration shall specify a brace-or-equal-initializer in which every initializer-clause that is an assignment-expression is a constant expression. [ Note: In both these cases, the member may appear in constant expressions. —end note ] The member shall still be defined in a namespace scope if it is odr-used (3.2) in the program and the namespace scope definition shall not contain an initializer.

Consider for example two programs

struct A { const static double x = 1.0; }; int main() { return 0; } struct A { constexpr static double x = 1.0; }; int main() { return 0; }

The first one will not compile while the second one will compile.

The same is valid for pointers

This program

struct A { static constexpr const char * myString = "myString"; }; int main() { return 0; }

will compile while this porgram

struct A { static const char * const myString = "myString"; }; int main() { return 0; }

will not compile.

2