C++ inline initialization of a static const char*

Question

Why I cannot initialize a static const char* in the header file? In my code I have in my Class header:

static const char* xml_ID_TAG;

and in the cpp:

const char* Class::xml_ID_TAG = "id";

The xml_ID_TAG variable contains the attribute string of an XML document. Since it's static, const, primitive type (char*), etc... I can't figure out why the compiler forbid to write something like:

static const char* xml_ID_TAG = "id";

I'm using MSVC2013 compiler, giving for the example above the error: "Error: a member with an in-class initializer must be const"

Only `static const` integral members can be defined inside a class body. This is to prevent ODR violations. — Simple, Jul 04 '14 at 10:47

Lightness Races in Orbit · Accepted Answer · 2019-06-14T10:28:31.780

Generally speaking you must define your static members in precisely one translation unit, and the language helps to enforce this by prohibiting you from writing an initialiser for such a member inside the surrounding class definition:

struct T
{
   static int x = 42;
   // ^ error: ISO C++ forbids in-class initialization of
   // non-const static member 'T::x'
};

However, a special exception is made for constants, for convenience:

struct T
{
   static const int x = 42;
   // ^ OK
};

Note that in most cases you still need to define the constant (in your .cpp file would be the best place):

const int T::x;

[C++11: 9.4.2/3]:] 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 assignment-expression 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.

Now, your member is not an int and even a const char* const is not of an "integral type":

struct T
{
   static const char* const str = "hi";
   // ^ error: 'constexpr' needed for in-class initialization of
   // static data member 'const char* const T::str' of non-integral type
};

but it is of a "literal type"; the upshot for you is that if you write it like this:

static constexpr const char* const xml_ID_TAG = "id";
//     ^^^^^^^^^             ^^^^^

you should be okay. (Note that you will still need to define it, until C++17.)

This probably makes more sense anyway: why would you want to change the pointer?

Nope, I tried the last solution (static constexpr const char * const foo="foo") and it gives me link errors in some cases if I don't define it. — galinette, Jun 14 '19 at 09:36
@galinette Yeah [you still need to define it](https://stackoverflow.com/a/51051942/560648) (though not since C++17!). Thanks, will clarify. — Lightness Races in Orbit, Jun 14 '19 at 10:27

score 5 · Answer 2 · answered Jul 04 '14 at 10:49

Because string literals (eg "id") are stored per file compilation unit. So if they are in a header file, there's a different instance stored for each source file that includes it. So your 'initialisation' is trying to store different values in a static variable for each compilation unit that #includes it..

score 4 · Answer 3 · edited May 23 '17 at 12:00

4

"Since it's [...] const" - Nope.

You need const char* const - assuming you're using C++11.

Otherwise you have to put the definition in the cpp file.

See: How to initialize a static const member in C++?

edited May 23 '17 at 12:00

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answered Jul 04 '14 at 10:45

Karoly Horvath

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C++ inline initialization of a static const char*

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