Search code examples
c++c++11destructorconstexpruser-defined-literals

Empty destructor vs literal destructor


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 ?


Solution

  • 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