填充和填充只是同一事物的两个方面:

包装或对齐是每个成员四舍五入的大小 填充是为匹配对齐而添加的额外空间

在mystruct_A中，假设默认对齐为4，则每个成员以4字节的倍数进行对齐。因为char的大小是1，所以a和c的填充是4 - 1 = 3个字节，而int b不需要填充，因为它已经是4个字节了。它对mystruct_B的工作方式相同。

填充将结构成员对齐到“自然”地址边界——例如，int成员将有偏移量，在32位平台上是mod(4) == 0。默认情况下，填充是开启的。它在你的第一个结构中插入以下“间隙”:

struct mystruct_A {
    char a;
    char gap_0[3]; /* inserted by compiler: for alignment of b */
    int b;
    char c;
    char gap_1[3]; /* -"-: for alignment of the whole struct in an array */
} x;

另一方面，打包可以防止编译器进行填充-这必须显式地请求-在GCC下，它是__attribute__((__packked__))，因此如下:

struct __attribute__((__packed__)) mystruct_A {
    char a;
    int b;
    char c;
};

会在32位架构上产生大小为6的结构。

不过需要注意的是，在允许未对齐内存访问的体系结构(如x86和amd64)上，未对齐内存访问速度较慢，并且在严格对齐的体系结构(如SPARC)上是明确禁止的。

变量存储在可以被其对齐方式(通常是大小)整除的任何地址上。所以，填充/填充不仅仅是为了结构。实际上，所有数据都有自己的对齐要求:

int main(void) {
    // We assume the `c` is stored as first byte of machine word
    // as a convenience! If the `c` was stored as a last byte of previous
    // word, there is no need to pad bytes before variable `i`
    // because `i` is automatically aligned in a new word.

    char      c;  // starts from any addresses divisible by 1(any addresses).
    char pad[3];  // not-used memory for `i` to start from its address.
    int32_t   i;  // starts from any addresses divisible by 4.

这类似于struct，但有一些区别。首先，我们可以说有两种填充——a)为了正确地从每个成员的地址开始，在成员之间插入一些字节。b)为了正确地从struct的地址启动下一个struct实例，将一些字节追加到每个struct:

// Example for rule 1 below.
struct st {
    char      c;  // starts from any addresses divisible by 4, not 1.
    char pad[3];  // not-used memory for `i` to start from its address.
    int32_t   i;  // starts from any addresses divisible by 4.
};

// Example for rule 2 below.
struct st {
    int32_t   i;  // starts from any addresses divisible by 4.
    char      c;  // starts from any addresses.
    char pad[3];  // not-used memory for next `st`(or anything that has same
                  // alignment requirement) to start from its own address.
};

The struct's first member always starts from any addresses divisible by struct's own alignment requirement which is determined by largest member's alignment requirement(here 4, alignment of int32_t). This is different with normal variables. The normal variables can start any addresses divisible by its alignment, but it is not the case for struct's first member. As you know, the address of a struct is the same as the address of its first member. There can be additional padded trailing bytes inside a struct, making next struct(or next element in an array of structs) starting from its own address. Think of struct st arr[2];. To make arr[1](arr[1]'s first member) starting from an address divisible by 4, we should append 3 bytes at the end of each struct.

这是我从《丢失的结构包装艺术》中学到的。

注意:可以通过_Alignof操作符来研究数据类型的对齐要求。同样，你也可以通过offsetof宏来获取结构中成员的偏移量。

Data structure alignment is the way data is arranged and accessed in computer memory. It consists of two separate but related issues: data alignment and data structure padding. When a modern computer reads from or writes to a memory address, it will do this in word sized chunks (e.g. 4 byte chunks on a 32-bit system) or larger. Data alignment means putting the data at a memory address equal to some multiple of the word size, which increases the system’s performance due to the way the CPU handles memory. To align the data, it may be necessary to insert some meaningless bytes between the end of the last data structure and the start of the next, which is data structure padding.

In order to align the data in memory, one or more empty bytes (addresses) are inserted (or left empty) between memory addresses which are allocated for other structure members while memory allocation. This concept is called structure padding. Architecture of a computer processor is such a way that it can read 1 word (4 byte in 32 bit processor) from memory at a time. To make use of this advantage of processor, data are always aligned as 4 bytes package which leads to insert empty addresses between other member’s address. Because of this structure padding concept in C, size of the structure is always not same as what we think.

只有当你告诉编译器显式地对结构进行打包时，才会进行结构打包。你看到的是填充。您的32位系统正在填充每个字段以字对齐。如果您告诉编译器打包结构，它们将分别为6和5字节。但是不要这样做。它不可移植，使编译器生成的代码更慢(有时甚至有bug)。

这些结构是填充的还是包装的?

它们填充。

最初想到的唯一可能是，如果char和int的大小相同，那么char/int/char结构的最小大小将不允许填充，int/char结构也是如此。

然而，这将要求sizeof(int)和sizeof(char)都为4(以获得12和8的大小)。由于sizeof(char)始终为1的标准保证了整个理论的崩溃。

如果char和int的宽度相同，那么大小将是1和1，而不是4和4。因此，为了得到12的大小，在最终字段之后必须有填充。

什么时候进行填充或包装?

只要编译器实现需要。编译器可以在字段之间和最后一个字段之后(但不能在第一个字段之前)插入填充。

这样做通常是为了性能，因为某些类型在特定边界上对齐时性能更好。甚至有一些架构会在你试图访问未对齐的数据时拒绝运行(即崩溃)(是的，我在看你，ARM)。

您通常可以使用特定于实现的特性(如#pragma pack)来控制打包/填充(这实际上是同一领域的两个极端)。即使您不能在特定的实现中这样做，您也可以在编译时检查代码以确保它满足您的需求(使用标准C特性，而不是特定于实现的东西)。

例如:

// C11 or better ...
#include <assert.h>
struct strA { char a; int  b; char c; } x;
struct strB { int  b; char a;         } y;
static_assert(sizeof(struct strA) == sizeof(char)*2 + sizeof(int), "No padding allowed");
static_assert(sizeof(struct strB) == sizeof(char)   + sizeof(int), "No padding allowed");

如果这些结构中有任何填充，类似这样的东西将拒绝编译。