CPP warning : "‘B’ has a base ‘A<(& str)>’ whose type uses the anonymous namespace" while using a template string

Here is the simplest code I can give:

// main.cpp
#include "test.h"

int main() { return 0; }

// test.h
#pragma once
#include <cstring>

const char str[] = "ABCD";

template <
    const char* str,
    std::size_t N
>
class A {};

class B : public A<str, strlen(str)> {};

// test.cpp
#include "test.h"

When I compile, I get twice the following warning:

test.h:13:7: warning: ‘B’ has a base ‘A<(& str), 4>’ whose type uses the anonymous namespace [-Wsubobject-linkage]
   13 | class B : public A<str, strlen(str)> {};
      |       ^

The reason why I get it twice is because test.h is included twice, once in main.cpp and once in test.cpp.

A similar question has already been answering in this post, but there was only a const char* and not a std::size_t too.

The answer of the above linked post was to use the extern keyword before const char str[] and define str in the test.cpp. But with std::size_t the compilation with the corrected code gives the following error:

test.h:13:31: error: ‘strlen(((const char*)(& str)))’ is not a constant expression
   13 | class B : public A<str, strlen(str)> {};
      |                         ~~~~~~^~~~~

Of course my code is more complicated than that and the main problem is that my object only need a const char*, the strlen(str) is done by another templated class far behind.

How can I do things to not have warnings or errors?

>Solution :

std::strlen() can’t be used in a constexpr context. You can use std::size(). Also you need to make str array inline for letting the compiler know str array size.

inline const char str[] = "ABCD";

lass B : public A<str, std::size(str) - 1> {};