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.

我知道这个问题很老了，这里的大多数答案都很好地解释了填充，但当我自己试图理解它时，我发现对正在发生的事情有一个“视觉”形象是有帮助的。

处理器以一定大小(字)的“块”读取内存。假设处理器字有8字节长。它将把内存看作一个8字节的大行构建块。每当它需要从内存中获取一些信息时，它就会到达其中一个块并获取它。

如上图所示，一个Char(1字节长)在哪里并不重要，因为它将在其中一个块中，只需要CPU处理1个字。

When we deal with data larger than one byte, like a 4 byte int or a 8 byte double, the way they are aligned in the memory makes a difference on how many words will have to be processed by the CPU. If 4-byte chunks are aligned in a way they always fit the inside of a block (memory address being a multiple of 4) only one word will have to be processed. Otherwise a chunk of 4-bytes could have part of itself on one block and part on another, requiring the processor to process 2 words to read this data.

这同样适用于8字节的double，只不过现在它必须在8的倍数内存地址中，以确保它始终在块中。

这里考虑的是8字节的字处理器，但这个概念也适用于其他大小的字。

填充通过填充这些数据之间的间隙来确保它们与这些块对齐，从而提高读取内存时的性能。

然而，正如其他人回答的那样，有时空间比性能本身更重要。也许您正在一台没有太多RAM的计算机上处理大量数据(可以使用交换空间，但速度要慢得多)。您可以在程序中排列变量，直到完成最少的填充(这在其他一些回答中得到了很好的例子)，但如果这还不够，您可以显式地禁用填充，这就是打包。

填充规则:

结构体的每个成员都应该位于能被其大小整除的地址。 填充在元素之间或结构的末尾插入，以确保满足此规则。这样做是为了硬件更容易和更有效地访问总线。 结构体末尾的填充是根据结构体最大成员的大小决定的。

规则二: 考虑下面的结构，

如果我们要为这个结构体创建一个数组(包含2个结构体)， 结束时不需要填充:

因此，struct的大小= 8字节

假设我们要创建另一个结构体，如下所示:

如果我们要创建这个结构体的数组， 最后需要填充的字节数有两种可能。

A.如果我们在末尾添加3个字节，并将其对齐为int而不是Long:

B.如果我们在末尾添加7个字节并将其对齐为Long:

第二个数组的起始地址是8(i)的倍数。e 24)。 struct的大小= 24字节

因此，通过将结构体的下一个数组的起始地址对齐为最大成员(i。E如果我们要创建这个结构体的数组，第二个数组的第一个地址必须从一个地址开始，该地址必须是该结构体最大成员的倍数。这里是24(3 * 8))，我们可以计算出最后所需的填充字节数。

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。

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)。

这件事没有但是!想要掌握这门学科必须做到以下几点:

细读埃里克·s·雷蒙德所著的《丢失的结构包装艺术》 看一下Eric的代码示例 最后但并非最不重要的是，不要忘记下面关于填充的规则，即结构体的对齐方式与最大类型的对齐方式一致 要求。