Question

Consider the following code:

#include <iostream>

class Test
{
    public:
        constexpr Test(const int x) : _x(x) {}
        constexpr int get() const {return _x;}
        ~Test() {} // HERE
    protected:
        const int _x;
};

int main()
{
    static constexpr Test test(5);
    return 0;
}

If I remove the line HERE the code compiles well, but if I define an empty destructor, it results to a compilation error saying that Test is non-literal.

Why and what is the difference between an empty destructor and no destructor at all ?

EDIT: Another related question : if empty and literal destructors are different how to define a protected literal destructor ?

Answer 1

Quotes from n3376

7.1.5/9

A constexpr specifier used in an object declaration declares the object as const. Such an object shall have
literal type and shall be initialized. If it is initialized by a constructor call, that call shall be a constant
expression

3.9/10

A type is a literal type if:

it has a trivial destructor…

12.4/5

A destructor is trivial if it is not user-provided and if:

— the destructor is not virtual,

— all of the direct base classes of its class have trivial destructors, and

— for all of the non-static data members of its class that are of class type (or array thereof), each such
class has a trivial destructor.

Otherwise, the destructor is non-trivial.

clang diagnostic is really more informative:

error: constexpr variable cannot have non-literal type 'const C'

'C' is not literal because it has a user-provided destructor