C++ – Are All constexpr Variables Implicitly Inline?

c++c++17inline-variablelinkage

I was playing around with auto template parameters and I was surprised that this code didn't compiled:

constexpr auto bar = 2;

template<auto& T>
struct Foo {
    auto operator()() const { return T; }
};

int main() {
    Foo<bar> b;
    b();
}

Visual Studio 15.7 (preview 4) spit out these errors:

error C2970: 'Foo': template parameter 'T': 'bar': an expression involving objects with internal linkage cannot be used as a non-type argument
 note: see declaration of 'Foo'
 note: see declaration of 'bar'
error C2440: 'specialization': cannot convert from 'int' to 'int &'
 note: see reference to class template instantiation 'Foo<0>' being compiled
error C2973: 'Foo': invalid template argument 'int'
 note: see declaration of 'Foo'

Then, after adding inline, the error got away!

constexpr inline auto bar = 2;

I thought constexpr variables were implicitly inline. Also, how does that affect the linkage of my variable bar?

Best Answer

Are all constexpr variable implicitly inline?

No. Only constexpr functions and constexpr static data members are implicitly inline ([dcl.constexpr]/1).

Also, how does that affect the linkage of my variable bar?

A constexpr variable is const ([dcl.constexpr]/9). A non-inline const variable that is not explicitly declared extern has internal linkage ([basic.link]/3).

Related Solutions

C++11 – Does constexpr Imply Inline?

Yes ([dcl.constexpr], §7.1.5/2 in the C++11 standard): "constexpr functions and constexpr constructors are implicitly inline (7.1.2)."

Note, however, that the inline specifier really has very little (if any) effect upon whether a compiler is likely to expand a function inline or not. It does, however, affect the one definition rule, and from that perspective, the compiler is required to follow the same rules for a constexpr function as an inline function.

I should also add that regardless of constexpr implying inline, the rules for constexpr functions in C++11 required them to be simple enough that they were often good candidates for inline expansion (the primary exception being those that are recursive). Since then, however, the rules have gotten progressively looser, so constexpr can be applied to substantially larger, more complex functions.

C++ Standard – Namespace-Scoped Constexpr Variables and Internal Linkage

Yes, constexpr on an object declaration means that the object is const. See [dcl.constexpr]/9. And yes, that means that kSomeString in your example has internal linkage.

The species of ODR violation we are talking about here is not the definition of kSomeString itself, but other definitions that attempt to use it. And there's a problem precisely because of the internal linkage. Consider:

void f(const std::string_view &);

inline void g() { 
    f(foo::kSomeString); 
}

This is an ODR violation if included in multiple translation units, essentially because the definition of g in each translation unit references a different object.

Best Answer

Related Solutions

C++11 – Does constexpr Imply Inline?

C++ Standard – Namespace-Scoped Constexpr Variables and Internal Linkage

Related Question