std::aligned_storage 的分配目标（堆栈还是堆？）
我一直在尝试了解 TR1 添加的内容（称为aligned_storage）。在阅读以下文档N2165时， N3190 和 N2140 我一生都看不到语句，其中清楚地描述了正在使用的内存的堆栈或堆性质。
我查看了 msvc2010、boost 和 gcc 提供的实现，它们都提供了一个基于堆栈的解决方案，以联合的使用为中心。
简而言之：
aligned_storage实现使用的内存类型（堆栈或堆）是否已定义，或者它始终是基于堆栈的？
以及，定义/确定该内容的具体文档是什么？
注意：在MSVC10中，以下是aligned_storage类型的定义，在这种情况下，如果aligned_storage是自动变量，则在堆栈上创建data(_Val,_Pad)：
template<class _Ty, size_t _Len>
union _Align_type
{
// union with size _Len bytes and alignment of _Ty
_Ty _Val;
char _Pad[_Len];
};
注意：这是NOT 一个小问题。在发布答案之前，请尝试理解该问题。
I've been trying to get my head around the TR1 addition known as aligned_storage. Whilst reading the following documents N2165, N3190 and N2140 I can't for the life of me see a statement where it clearly describes stack or heap nature of the memory being used.
I've had a look at the implementation provided by msvc2010, boost and gcc they all provide a stack based solution centered around the use of a union.
In short:
Is the memory type (stack or heap) used by aligned_storage implementation defined or is it always meant to be stack based?
and, What the is the specific document that defines/determines that?
Note: In MSVC10, the following is the definition of the type of aligned_storage, in this case if the aligned_storage is an auto variable the data(_Val,_Pad) is created on the stack:
template<class _Ty, size_t _Len>
union _Align_type
{
// union with size _Len bytes and alignment of _Ty
_Ty _Val;
char _Pad[_Len];
};
Note: This is NOT a trivial question. Please try and understand the question before posting an answer.
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std::aligned_storage只是声明一个成员 typedef (type)。（这来自最新的 C++0x 草案 N3225，20.7.6.6 表 53，但 TR1 规范中的语言 N1836 实际上是相同的，除了在 C++0x 中
Align< /code> 模板参数的默认参数是最大对齐值。）std::aligned_storage本身不分配任何内存。您可以创建std::aligned_storage::type 类型的对象，并将该对象重新解释为满足上述要求的任何类型的对象。std::aligned_storage<Len, Align>just declares a member typedef (type).(This is from the latest C++0x draft, N3225, 20.7.6.6 Table 53, but the language in the TR1 specification, N1836, is effectively the same except that in C++0x the
Aligntemplate parameter has as its default argument the maximum alignment value.)std::aligned_storagedoesn't allocate any memory itself. You can create an object of typestd::aligned_storage<Len, Align>::typeand reinterpret that object as an object of any type that meets the requirements stated above.您通常不需要对齐堆上的内容，因为任何分配（
new/malloc）都会在与任何类型对齐的地址处返回内存。You typically don't need to align stuff on the heap since any allocation (
new/malloc) returns memory at an address which is aligned to any type.