是一个结构的大小要求是该结构的调整的整数倍？ [英] Is the size of a struct required to be an exact multiple of the alignment of that struct?

问题描述

再一次，我质问一个长期的信念。

Once again, I'm questioning a longstanding belief.

直到今天，我认为以下结构的调整通常是4和大小通常是5 ...

Until today, I believed that the alignment of the following struct would normally be 4 and the size would normally be 5...

struct example
{
  int   m_Assume_32_Bits;
  char  m_Assume_8_Bit_Bytes;
};

这个假设，因为，我有一个使用offsetof，来确定数组中的两个相邻项目之间以字节为单位的距离数据结构code。今天，我发现它不应该，不明白为什么我没有过错误的话，codeD了单元测试的一些旧的code，它是用的sizeof - 和测试通过传递让我吃惊

Because of this assumption, I have data structure code that uses offsetof to determine the distance in bytes between two adjacent items in an array. Today, I spotted some old code that was using sizeof where it shouldn't, couldn't understand why I hadn't had bugs from it, coded up a unit test - and the test surprised me by passing.

调查有点表明，我用于测试（类似于上面的结构）sizeof的类型是对齐的整数倍 - 即8个字节。它的最终成员后，有填充。这就是为什么我从没有想到这...一个例子

A bit of investigation showed that the sizeof the type I used for the test (similar to the struct above) was an exact multiple of the alignment - ie 8 bytes. It had padding after the final member. Here is an example of why I never expected this...

struct example2
{
  example m_Example;
  char    m_Why_Cant_This_Be_At_Offset_6_Bytes;
};

谷歌搜索的结果显示一点例子，说清楚，最后的成员后，该填充是允许的 - 例如：<一href="http://en.wikipedia.org/wiki/Data_structure_alignment#Data_structure_padding">http://en.wikipedia.org/wiki/Data_structure_alignment#Data_structure_padding （下称或在该结构的端部位）。

A bit of Googling showed examples that make it clear that this padding after the final member is allowed - for example http://en.wikipedia.org/wiki/Data_structure_alignment#Data_structure_padding (the "or at the end of the structure" bit).

这是一个有点尴尬，因为我最近发布此评论 - 使用结构填充（我的第一个注释，这个问题的答案）。

This is a bit embarrassing, as I recently posted this comment - Use of struct padding (my first comment to that answer).

我似乎无法确定的是这个填充到对准的整数倍是否由C ++标准的保证，还是它只是东西是允许的，一些（但可能不是全部）编译器做的。

What I can't seem to determine is whether this padding to an exact multiple of the alignment is guaranteed by the C++ standard, or whether it is just something that is permitted and that some (but maybe not all) compilers do.

所以 - 是按照C ++标准是该结构的调整的整数倍需要结构的大小

So - is the size of a struct required to be an exact multiple of the alignment of that struct according to the C++ standard?

如果C标准使得不同的保证，我很感兴趣的是太多，但重点是C ++。

If the C standard makes different guarantees, I'm interested in that too, but the focus is on C++.

推荐答案

的一个定义<打击>调整大小：

的<击>调整的结构体的大小是由一个元素偏移到下一个元素，当你有一个结构的数组。

The alignment size of a struct is the offset from one element to the next element when you have an array of that struct.

由于其性质，如果你有一个结构中有两个元素的数组，然后都需要有一致的成员，这样就意味着是的，尺寸必须对齐的倍数。 （我不知道是否任何标准明确强制执行这一点，但由于结构的大小和对齐方式不依赖于结构是否单独或数组里面，同样的规则适用于，所以它不能真正是任何其他方式。）

By its nature, if you have an array of a struct with two elements, then both need to have aligned members, so that means that yes, the size has to be a multiple of the alignment. (I'm not sure if any standard explicitly enforce this, but because the size and alignment of a struct don't depend on whether the struct is alone or inside an array, the same rules apply to both, so it can't really be any other way.)

这篇关于是一个结构的大小要求是该结构的调整的整数倍？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！