g++-4.8.1 thinks that an explicitly-declared destructor with no exception specification is always noexcept(true)

Question

Consider the following program:

#include <type_traits>

struct Thrower
{
    ~Thrower() noexcept(false) { throw 1; }
};

struct Implicit
{
    Thrower t;
};
static_assert(!std::is_nothrow_destructible<Implicit>::value, "Implicit");

struct Explicit
{
    ~Explicit() {}

    Thrower t;
};
static_assert(!std::is_nothrow_destructible<Explicit>::value, "Explicit");

With g++-4.8.1, there is a static assertion failure on Explicit -- it seems to think that ~Explicit() is noexcept. This does not match my expectations. According to §12.4.3:

A declaration of a destructor that does not have an exception-specification is implicitly considered to have the same exception-specification as an implicit declaration

The funny thing here is the check of Implicit seems to be behaving according to my interpretation of §15.4.14 (through §12.4.7).

...If f is an...destructor...it's implicit exception-specification specifies...f has the exception-specification noexcept(true) if every function it directly invokes allows no exceptions.

g++-4.7 lacks is_nothrow_destructable, I wrote my own to check the behavior in 4.7. The program seems to compile perfectly fine. I reserve the right for this to be completely wrong and the source of my confusion:

template <typename T>
struct is_nothrow_destructible
{
    static constexpr bool value = noexcept(std::declval<T>().~T());
};

TL;DR: Why does g++-4.8.1 think that an explicitly-declared destructor with no exception specification is always noexcept(true)?

Update: I opened a bug on this: 57645. If you really need to work around this issue, you can add an exception specification to the destructor (like Thrower has in the example).

And that's how Andy Prowl wrote an answer with 1000+ upvotes. — , Jun 18 '13 at 20:06
@H2CO3: lol, no, I just think this is a compiler bug and I have zero chance to say anything relevant about it ;) — Andy Prowl, Jun 18 '13 at 20:33

score 7 · Accepted Answer · answered Jun 18 '13 at 20:34

TL;DR: Why does g++-4.8.1 think that an explicitly-declared destructor with no exception specification is always noexcept(true)?

Because it has a bug?

Your interpretation of the standard is correct, and Clang implements it correctly (the assert doesn't fire).

f has the exception-specification noexcept(true) if every function it directly invokes allows no exceptions.

A destructor directly invokes the destructor of all subobjects:

§12.4 [class.dtor] p8:

After executing the body of the destructor and destroying any automatic objects allocated within the body, a destructor for class X calls the destructors for X’s direct non-variant non-static data members, [...].

Oh yeah, please file a bugreport [over at GCC's bugtracker](http://gcc.gnu.org/bugzilla/). — Xeo, Jun 18 '13 at 20:54
I was hoping I was wrong instead of the compiler...it is a much more comfortable thing. I opened a bug on it. — Travis Gockel, Jun 19 '13 at 05:09

g++-4.8.1 thinks that an explicitly-declared destructor with no exception specification is always noexcept(true)

1 Answers1