Are all constexpr variable implicitly inline?

问题

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?

回答1:

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).

来源：https://stackoverflow.com/questions/49913011/are-all-constexpr-variable-implicitly-inline