什么时候应该使用工会?我们为什么需要它们?


当前回答

在C的早期版本中,所有结构声明都共享一组公共字段。考虑到:

struct x {int x_mode; int q; float x_f};
struct y {int y_mode; int q; int y_l};
struct z {int z_mode; char name[20];};

a compiler would essentially produce a table of structures' sizes (and possibly alignments), and a separate table of structures' members' names, types, and offsets. The compiler didn't keep track of which members belonged to which structures, and would allow two structures to have a member with the same name only if the type and offset matched (as with member q of struct x and struct y). If p was a pointer to any structure type, p->q would add the offset of "q" to pointer p and fetch an "int" from the resulting address.

Given the above semantics, it was possible to write a function that could perform some useful operations on multiple kinds of structure interchangeably, provided that all the fields used by the function lined up with useful fields within the structures in question. This was a useful feature, and changing C to validate members used for structure access against the types of the structures in question would have meant losing it in the absence of a means of having a structure that can contain multiple named fields at the same address. Adding "union" types to C helped fill that gap somewhat (though not, IMHO, as well as it should have been).

An essential part of unions' ability to fill that gap was the fact that a pointer to a union member could be converted into a pointer to any union containing that member, and a pointer to any union could be converted to a pointer to any member. While the C89 Standard didn't expressly say that casting a T* directly to a U* was equivalent to casting it to a pointer to any union type containing both T and U, and then casting that to U*, no defined behavior of the latter cast sequence would be affected by the union type used, and the Standard didn't specify any contrary semantics for a direct cast from T to U. Further, in cases where a function received a pointer of unknown origin, the behavior of writing an object via T*, converting the T* to a U*, and then reading the object via U* would be equivalent to writing a union via member of type T and reading as type U, which would be standard-defined in a few cases (e.g. when accessing Common Initial Sequence members) and Implementation-Defined (rather than Undefined) for the rest. While it was rare for programs to exploit the CIS guarantees with actual objects of union type, it was far more common to exploit the fact that pointers to objects of unknown origin had to behave like pointers to union members and have the behavioral guarantees associated therewith.

其他回答

当你有一个函数,你返回的值可以不同,这取决于函数做了什么,使用联合。

联合允许互斥的数据成员共享相同的内存。当内存比较稀缺时,例如在嵌入式系统中,这是非常重要的。

示例如下:

union {
   int a;
   int b;
   int c;
} myUnion;

这个联合将占用一个int值的空间,而不是3个独立的int值。如果用户设置了a的值,然后设置了b的值,它将覆盖a的值,因为它们都共享相同的内存位置。

联合在嵌入式编程或需要直接访问硬件/内存的情况下特别有用。这里有一个简单的例子:

typedef union
{
    struct {
        unsigned char byte1;
        unsigned char byte2;
        unsigned char byte3;
        unsigned char byte4;
    } bytes;
    unsigned int dword;
} HW_Register;
HW_Register reg;

然后,您可以按如下方式访问reg:

reg.dword = 0x12345678;
reg.bytes.byte3 = 4;

字节顺序和处理器体系结构当然很重要。

另一个有用的特性是位修饰符:

typedef union
{
    struct {
        unsigned char b1:1;
        unsigned char b2:1;
        unsigned char b3:1;
        unsigned char b4:1;
        unsigned char reserved:4;
    } bits;
    unsigned char byte;
} HW_RegisterB;
HW_RegisterB reg;

使用这段代码,您可以直接访问寄存器/内存地址中的单个位:

x = reg.bits.b2;

在学校里,我是这样使用联合的:

typedef union
{
  unsigned char color[4];
  int       new_color;
}       u_color;

我用它来更容易地处理颜色,而不是使用>>和<<操作符,我只需要遍历我的char数组的不同索引。

一个简单而有用的例子是....

想象一下:

你有一个uint32_t数组[2],想要访问字节链的第3个和第4个字节。 你可以做*((uint16_t*) &数组[1])。 但遗憾的是,这打破了严格的混叠规则!

但是已知的编译器允许你做以下事情:

union un
{
    uint16_t array16[4];
    uint32_t array32[2];
}

严格来说,这仍然是违反规则的。但是所有已知的标准都支持这种用法。