在C-结构的内存对齐 [英] Memory alignment in C-structs

问题描述

我工作的32位计算机上，所以我想，内存对齐应该是4个字节。说我有结构：

I'm working on the 32-bit machine, so I suppose that memory alignment should be 4 bytes. Say I have struct:

typedef struct {
    unsigned short v1;
    unsigned short v2;
    unsigned short v3;
} myStruct;

真正的大小为6个字节，我想这对齐大小应为8，但的sizeof（MYSTRUCT）返回我6。

但是，如果我写的：

typedef struct {
    unsigned short v1;
    unsigned short v2;
    unsigned short v3;
    int i;
} myStruct;

真正的大小为10字节，排列为12，这个时间的sizeof（MYSTRUCT）== 12 。

有人可以解释的区别是什么？

Can somebody explain what is the difference?

推荐答案

至少在大多数机器上，一个类型只会对准一样大类型本身[编辑边界：你真的不能要求任何更多对齐比，因为你必须要能够创建数组，你不能插入填充到一个数组。在您的实现，短显然是2个字节，而 INT 4个字节。

At least on most machines, a type is only ever aligned to a boundary as large as the type itself . On your implementation, short is apparently 2 bytes, and int 4 bytes.

这意味着你的第一个结构对齐到2字节边界。由于所有的成员都是2个字节每人，不填充插入它们之间。

That means your first struct is aligned to a 2-byte boundary. Since all the members are 2 bytes apiece, no padding is inserted between them.

第二包含一个4字节的项目，其被对准以4个字节的边界。由于它是由6个字节pceded $ P $，2字节的填充插入 V3 和 I ，给予6个字节在短 s的数据，两个字节的填充物，并且总共4个字节的 INT 数据的12。

The second contains a 4-byte item, which gets aligned to a 4-byte boundary. Since it's preceded by 6 bytes, 2 bytes of padding is inserted between v3 and i, giving 6 bytes of data in the shorts, two bytes of padding, and 4 more bytes of data in the int for a total of 12.

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