C结构中的压缩位字段 - GCC

问题描述

我正在使用linux上的c结构。
我开始使用位字段和packed属性，并且遇到了一个奇怪的行为：

$ $ $ $ $ $ $ $ $ struct $
{
int a：12;
int b：32;
int c：4;
} __属性__（（packed））;

struct t2
{
int a：12;
int b;
int c：4;
} __属性__（（packed））;

void main（）
{
printf（％d\\\
，sizeof（t1））; //输出 - 6
printf（％d \ n，sizeof（t2））; //输出 - 7
}

两个结构如何 - 完全相同 - 需要不同的字节数？

解决方案

您的结构不是完全相同。你的第一个有三个连续的位域，第二个有一个位域，一个（非位域）int，然后是第二个位域。

这是有意义的：连续（非零宽度）位字段合并到单个存储位置中，而位字段后跟非位字段是不同的存储位置。

>

您的第一个结构有一个内存位置，第二个结构有三个。您可以在第二个结构中取得 b 成员的地址，而不是第一个结构中的地址。对 b 成员的访问不会与访问 a 或 c 在你的第二个结构中，但它们在你的第一个结构中。

紧接着有一个非位字段（或一个零长度的位字段）一个位域成员在一个敏感中关闭它，接下来将是一个不同的/独立的内存位置/对象。编译器无法像在第一个结构中一样打包您的 b 成员。

I am working with structs in c on linux.I started using bit fields and the "packed" attribute and I came across a wierd behavior:

struct t1
{
    int a:12;
    int b:32;
    int c:4;
}__attribute__((packed));

struct t2
{
    int a:12;
    int b;
    int c:4;
}__attribute__((packed));

void main()
{
    printf("%d\n",sizeof(t1)); //output - 6
    printf("%d\n",sizeof(t2)); //output - 7
}

How come both structures - that are exactly the same - take diffrent number of bytes?

解决方案

Your structures are not "exactly the same". Your first one has three consecutive bit-fields, the second has one bit-field, an (non bit-field) int, and then a second bit-field.

This is significant: consecutive (non-zero width) bit-fields are merged into a single memory location, while a bit-field followed by a non-bit-field are distinct memory locations.

Your first structure has a single memory location, your second has three. You can take the address of the b member in your second struct, not in your first. Accesses to the b member don't race with accesses the a or c in your second struct, but they do in your first.

Having a non-bit-field (or a zero-length bit-field) right after a bit-field member "closes" it in a sens, what follow will be a different/independent memory location/object. The compiler cannot "pack" your b member inside the bit-field like it does in the first struct.

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