是的sizeof（T）==的sizeof（int）的？ [英] Is sizeof(T) == sizeof(int)?

问题描述

我一直在钻研的标准草案，似乎无法找到我要找的。

如果我有一个标准布局类型。

 结构T {
   无符号的手柄;
};
 

后来我知道， reinter pret_cast＆LT;无符号*＆GT;（＆amp; T公司）==＆安培; t.handle 一些 T（T）;

我们的目标是创建一些矢量＆lt; T＆GT; v 并通过＆放大器;五[0] C函数期望一个指向无符号整数数组

因此​​，没有标准的定义的sizeof（T）==的sizeof（无符号）及这是否意味着 T的数组将有相同布局的数组无符号？

问题解决了一个非常类似的话题，我米询问其中两个数据成员和类是标准布局的特定情况下，与数据成员是一个基本的类型。

我读过一些段落，似乎暗示的也许的可能是真实的，但没有什么击中了要害。例如：

第9.2.17

  

两个标准布局结构（第9条）类型是布局兼容，如果
  它们具有相同数量的非静态数据成员和相应的
  非静态数据成员（按照声明顺序）具有布局兼容
  类型

这是并不完全符合我要找的，我不认为。

解决方案

您基本上都在问，给出：

 结构T {
    ü处理;
};
 

不管是保证的sizeof（T）==的sizeof（U）。不，不是。

在ISO C ++标准03第9.2 / 17说道：

  

一个指向POD结构对象，适当使用转换
   reinter pret_cast ，指出了其最初的成员​​（如该成员是
  位字段，然后在它所在的单元），反之亦然。

假设你有的结构T 的数组。该反之亦然部分意味着任何一个结构T T ::处理成员也必须是一个有效的地址的地址$ C>。现在，假设这些成员的类型的字符和你的说法是真实的。这意味着结构T 将被允许有一个未对齐的地址，这似乎是相当不可能的。标准通常设法不配合实现的手中以这样的方式。对于你所说的是真实的，该标准将不得不要求结构T 允许有不对齐的地址。而且它必须被允许用于所有结构，因为结构T 可能是一个前瞻性声明的，不透明的类型。

此外，部分9.2 / 17继续说：

  

[注：因此，有可能是一个POD结构对象中未命名的填充，而不是在它的开始，因为要达到相应的调整]

其中，采取了不同的方式，意味着不保证绝不会有填充

I've been poring over the draft standard and can't seem to find what I'm looking for.

If I have a standard-layout type

struct T {
   unsigned handle;
};

Then I know that reinterpret_cast<unsigned*>(&t) == &t.handle for some T t;

The goal is to create some vector<T> v and pass &v[0] to a C function that expects a pointer to an array of unsigned integers.

So, does the standard define sizeof(T) == sizeof(unsigned) and does that imply that an array of T would have the same layout as an array of unsigned?

While this question addresses a very similar topic, I'm asking about the specific case where both the data member and the class are standard layout, and the data member is a fundamental type.

I've read some paragraphs that seem to hint that maybe it might be true, but nothing that hits the nail on the head. For example:

§ 9.2.17

Two standard-layout struct (Clause 9) types are layout-compatible if they have the same number of non-static data members and corresponding non-static data members (in declaration order) have layout-compatible types

This isn't quite what I'm looking for, I don't think.

解决方案

You essentially are asking, given:

struct T {
    U handle;
};

whether it's guaranteed that sizeof(T) == sizeof(U). No, it is not.

Section 9.2/17 of the ISO C++03 standard says:

A pointer to a POD-struct object, suitably converted using a reinterpret_cast, points to its initial member (or if that member is a bit-field, then to the unit in which it resides) and vice versa.

Suppose you have an array of struct T. The vice versa part means that the address of any of the T::handle members must also be a valid address of a struct T. Now, suppose that these members are of type char and that your claim is true. This would mean that struct T would be allowed to have an unaligned address, which seems rather unlikely. The standard usually tries to not tie the hands of implementations in such a way. For your claim to be true, the standard would have to require that struct T be allowed to have unaligned addresses. And it would have to be allowed for all structures, because struct T could be a forward-declared, opaque type.

Furthermore, section 9.2/17 goes on to state:

[Note: There might therefore be unnamed padding within a POD-struct object, but not at its beginning, as necessary to achieve appropriate alignment.]

Which, taken a different way, means that there is no guarantee that there will never be padding.

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