On Tue, Oct 12, 2010 at 11:08 AM, Daniel Walker <daniel.j.walker@gmail.com> wrote:
True, that can be done, but boost::function is still default constructible and can still be cleared through assigning 0. So, you end up back to the exception safety question of how operator() behaves when the wrapper is empty. As you suggest, you could have the behavior depend on how it was constructed, so sometimes boost::function has a strong exception guarantee, sometimes it has a nothrow guarantee (and presumably, it's a no-op when empty).
Here is some code to make sure we're talking about the same thing: #include <new> using namespace std; extern void throw_exception(); template <class T> class function { public: void (*f)(); function():f(&throw_exception) { } function( nothrow_t ): f(0) { } void operator()() { f(); } }; int main() { function<int()> f(); //by default, you get link error. //function<int()> f(nothrow); //crash, but no link error. f(); } It would have been nicer if op() didn't throw to begin with but because that's how it has been, this should remain the default behavior. Emil Dotchevski Reverge Studios, Inc. http://www.revergestudios.com/reblog/index.php?n=ReCode