Access to constexpr variable inside lambda expression without capturing

Question

In the following example, I can access the constexpr variable x from inside the lambda y without explicitly capturing it. This is not possible if x is not declared as constexpr.

Are there special rules that apply to constexpr for capturing?

int foo(auto l) {
    // OK
    constexpr auto x = l();
    auto y = []{return x;};
    return y();

    // NOK
    // auto x2 = l();
    // auto y2 = []{ return x2; };
    // return y2();        
}

auto l2 = []{return 3;};

int main() {
    foo(l2);
}

Answer 1

Are there special rules that apply to constexpr for capturing/accessing?

Yes, constexpr variables could be read without capturing in lambda:

A lambda expression can read the value of a variable without capturing it if the variable

• has const non-volatile integral or enumeration type and has been initialized with a constant expression, or
• is constexpr and trivially copy constructible.