为什么一个结构中的内存对齐的变化？ [英] Why does the memory alignment change within a structure?

问题描述

在<一个href="http://stackoverflow.com/questions/27841898/why-does-my-structure-element-carry-padding-bytes">$p$pvious问题，我了解到，当有8字节对齐的结构被嵌入到其他结构与4-字节对齐，填充需要的 8字节对齐的结构之前。

明白了。

至少我认为我的理解。

借助 VS 2012文档说：

  

有关结构，工会和数组，对齐，要求其成员中最大的对准要求。

所以，如果我有一个结构是这样的：

  typedef结构s_inner {
  无符号长UL1;
  双DBL1;
  fourth_struct S4;
  无符号长UL2;
  INT I1;
} t_inner;
 

我期望的此结构的所有成员都将是8字节对齐，因为双 8字节对齐。

但我的内存转储显示我：

t_inner 从地址开始 1B8 ：

• 1B8 ：在无符号长是因为填充结构为8字节对齐
• 1C0 ：在双占用8个字节
• 1C8 ： fourth_struct 如下，它有4个字节对齐

到现在为止，一切都如预期。但是，现在的在 t_inner 对齐开关： 在地址 1E8 ，我希望找到刚无符号长这里，补齐4字节，因此，以下 INT 也排列在8个字节。但它好像对准现在因为改变unsigned long类型并的没有的进行填充字节。相反，下面的 INT 放置了4字节对齐。

为什么在 t_inner 对齐开关？其中的规则在这里应用？

解决方案

  

我期望这个结构的所有成员都将是自8字节对齐   双是8字节>对齐。

哦，不。每个键入的都有自己的定位的在的结构体，而的结构本身的有定位即是其中的最大的它的内容比对。

 对齐。 typedef结构s_inner {
      无符号长UL1; // 4对齐
      双DBL1; // 8对齐（需要4填充在此之前）
      fourth_struct S4; // 4对齐尺寸32
      无符号长UL2; // 4对齐（无填充）
      INT I1; // 4对齐（无填充）
      //无需为结构填充为8字节的整数倍
    } t_inner;
 

该结构本身具有的，因为双对齐8。

In a previous question, I learned that when a structure that has 8-byte alignment gets embedded into another structure with 4-byte alignment, a padding is needed before the 8-byte aligned structure.

Understood.

At least I thought that I understood.

The VS 2012 docs say:

For structures, unions, and arrays, the alignment-requirement is the largest alignment-requirement of its members.

So, if I have a structure like this:

typedef struct s_inner {
  unsigned long ul1;
  double        dbl1;
  fourth_struct s4;
  unsigned long ul2;
  int           i1;
} t_inner;

I'd expect all members of this structure would be 8-byte aligned since the double is 8-byte aligned.

But my memory dump shows me this:

t_inner starts from address 1B8:

• 1B8: the unsigned long is padded because the structure is 8-byte aligned
• 1C0: The double consumes 8 bytes
• 1C8: fourth_struct follows, it has 4-byte alignment

Until now, everything is as expected. But now the alignment switches inside t_inner: On address 1E8, I'd expect to find just the unsigned long here, padded with 4 bytes so that the following int is also aligned on 8 bytes. But it seems as if the alignment has now changed since the unsigned long does not carry padding bytes. Instead, the following int is placed with a 4-byte alignment.

Why does the alignment switch inside t_inner? Which rule is applied here?

解决方案

I'd expect all members of this structure would be 8-byte aligned since the double is 8-byte > aligned.

Well, no. Each type has its own alignment inside the struct, and the struct itself has an alignment that is the greatest of the alignments of it's content.

aligned.    typedef struct s_inner {
      unsigned long ul1;        // 4-aligned
      double        dbl1;       // 8-aligned  (need 4 padding before this)
      fourth_struct s4;         // 4 aligned  size 32
      unsigned long ul2;        // 4 aligned  (no padding)
      int           i1;         // 4 aligned  (no padding)
      // no padding needed as struct is a multiple of 8 bytes
    } t_inner; 

The struct itself has align 8 because of the double.

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