Is move assignment via destruct+move construct safe?

问题

Here's a very easy way to define move assignment for most any class with a move constructor:

class Foo {
public:
  Foo(Foo&& foo);                     // you still have to write this one
  Foo& operator=(Foo&& foo) {
    if (this != &foo) {               // avoid destructing the only copy
      this->~Foo();                   // call your own destructor
      new (this) Foo(std::move(foo)); // call move constructor via placement new
    }
    return *this;
  }
  // ...
};

Is this sequence of calling your own destructor followed by placement new on the this pointer safe in standard C++11?

回答1:

Only if you never, ever derive a type from this class. If you do, this will turn the object into a monstrosity. It's unfortunate that the standard uses this as an example in explaining object lifetimes. It's a really bad thing to do in real-world code.

回答2:

Technically, the source code is safe in this tiny example. But the reality is that if you ever even look at Foo funny, you will invoke UB. It's so hideously unsafe that it's completely not worth it. Just use swap like everybody else- there's a reason for it and it's because that's the right choice. Also, self-assignment-checking is bad.

来源：https://stackoverflow.com/questions/13092240/is-move-assignment-via-destructmove-construct-safe