鉴于任何POD类型,建议采取类似行动:
any_pod* p = new any_pod[n];
for (std::size_t i = 0; i < n; ++i)
new (&p[i].member) other_pod(whatever);
鉴于这个问题略为主观,我永远不会提出这样的法典。 无论它是否采取不明确的行为,它都难以读到<>>>>,并将要求任何未来的维护者花更多的时间(可能很重要)来 gr清你重新做的事情。
如果您需要不同的类型能力,要么使用<代码>boost:variant,要么boost:any,视您的需要而定。
If you just want to take the data from one class and copy/assign it to another, that s what converting constructors and converting assignment operators are for.
你在已经建造的物体上建树,再次违反了销毁的语文保证,因此,我不会这样做,不管物体是裁军部。
我很想知道,你是否试图问一下什么稍有不同:
any_pod* p = reinterpret_cast<any_pod*>(malloc(n * sizeof(any_pod[n]));
for (std::size_t i = 0; i < n; ++i)
new (&p[i].member) other_pod(whatever);
在此情况下,你不重建,你只是把建筑在一个原始记忆中。 在这种情况下,安置是适当的。 (尽管很罕见的是,你不得不照此写法典。) 在执行诸如:主人”之类的事情时,你可以做些什么。 你们必须非常小心,在最后一枚正好被拆除后释放阵列。
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