Destroy and then construct new object using the same variable

Sometimes it's nice to start over. In C++ I can employ this following simple manoeuvre:

{

    T x(31, Blue, false);

    x.~T();                        // enough with the old x

    ::new (&x) T(22, Brown, true); // in with the new!

    // ...
}

At the end of the scope, the destructor will run once again and all seems well. (Let's also say T is a bit special and doesn't like being assigned, let alone swapped.) But something tells me that it's not always without risk to destroy everything and try again. Is there a possible catch with this approach?

Solution 1:

I think the only way to make this really safe to use is to require the called constructor to be noexcept, for example by adding a static_assert:

static_assert(noexcept(T(22, Brown, true)), "The constructor must be noexcept for inplace reconstruction");
T x(31, Blue, false);
x.~T();
::new (&x) T(22, Brown, true);

Of course this will only work for C++11.

Solution 2:

If T's constructor throws on the second construction, you got a problem. If you like brute-force approaches, check this:

T x(31, Blue, false);
x.~T();
const volatile bool _ = true;
for(;_;){
  try{
    ::new (&x) T(22, Brown, true);
    break; // finally!
  }catch(...){
    continue; // until it works, dammit!
  }
}

It even provides the strong exception guarantee!

On a more serious note, it's like stepping on a landmine, knowing it will go off if you move your foot...

And there actually is a way around the undefined behaviour of the double destruction here:

#include <cstdlib>

T x(31, Blue, false);
x.~T();
try{
  ::new (&x) T(22, Brown, true);
}catch(...){
  std::exit(1); // doesn't call destructors of automatic objects
}