C 结构中的内存对齐
我在 32 位机器上工作，所以我想内存对齐应该是 4 个字节.假设我有这个结构:
I'm working on a 32-bit machine, so I suppose that the memory alignment should be 4 bytes. Say I have this struct:
typedef struct {
unsigned short v1;
unsigned short v2;
unsigned short v3;
} myStruct;
普通添加的大小是 6 个字节，我想对齐的大小应该是 8，但是 sizeof(myStruct) 返回我 6.
The plain added size is 6 bytes, and I suppose that the aligned size should be 8, but sizeof(myStruct) returns me 6.
但是如果我写:
typedef struct {
unsigned short v1;
unsigned short v2;
unsigned short v3;
int i;
} myStruct;
普通添加的大小是10个字节，对齐的大小应该是12，这次sizeof(myStruct) == 12.
the plain added size is 10 bytes, aligned size shall be 12, and this time sizeof(myStruct) == 12.
有人能解释一下有什么区别吗?
Can somebody explain what is the difference?
推荐答案
至少在大多数机器上，类型只与类型本身一样大的边界对齐.在您的实现中，short 显然是 2 个字节，而 int 是 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.
这意味着您的第一个结构与 2 字节边界对齐.由于所有成员各占2个字节，因此它们之间没有插入填充.
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.
第二个包含一个 4 字节的项目，它与 4 字节的边界对齐.由于前面是6个字节，所以在v3和i之间插入了2个字节的padding，在shorts中给出了6个字节的数据，两个填充字节，以及 int 中另外 4 个字节的数据，总共 12 个.
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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