使用相同的变量销毁并构造新对象

问题描述

有时很好，重新开始。在C ++中，我可以使用以下简单的操作：

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!

    // ...
}

的范围，析构函数将再次运行，一切似乎很好。 （让我们还说 T 有点特别，不喜欢被分配，更不用说交换了。）但是有些东西告诉我，毁灭一切并不总是没有风险，再次。使用这种方法有可能捕获吗？

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?

推荐答案

我认为唯一的方法是让这个真正安全的使用是要求称为构造函数 noexcept ，例如通过添加 static_assert ：

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);

当然，这只适用于C ++ 11。

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

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