我可以为一个 union 成员分配一个值并从另一个 union 成员那里读取相同的值吗？

Question

基本上，我有一个

struct foo {
        /* variable denoting active member of union */
        enum whichmember w;
        union {
                struct some_struct my_struct;
                struct some_struct2 my_struct2;
                struct some_struct3 my_struct3;
                /* let's say that my_struct is the largest member */
        };
};

main()
{
        /*...*/
        /* earlier in main, we get some struct foo d with an */
        /* unknown union assignment; d.w is correct, however */
        struct foo f;
        f.my_struct = d.my_struct; /* mystruct isn't necessarily the */
                                /* active member, but is the biggest */
        f.w = d.w;
        /* code that determines which member is active through f.w */
        /* ... */
        /* we then access the *correct* member that we just found */
        /* say, f.my_struct3 */

        f.my_struct3.some_member_not_in_mystruct = /* something */;
}

Accessing C union members via pointers似乎说通过指针访问成员是可以的。看评论。

但我的问题涉及直接访问它们。基本上，如果我将我需要的所有信息写入 union 的最大成员并手动跟踪类型，那么每次访问手动指定的成员仍然会产生正确的信息吗？

Answer 1

我注意到问题中的代码使用了匿名 union ，这意味着它必须为 C11 编写；匿名 union 不是 C90 或 C99 的一部分。
ISO/IEC 9899:2011 ，当前的 C11 标准，有这样的说法:

§6.5.2.3 Structure and union members

¶3 A postfix expression followed by the . operator and an identifier designates a member of a structure or union object. The value is that of the named member,⁹⁵⁾ and is an lvalue if the first expression is an lvalue. If the first expression has qualified type, the result has the so-qualified version of the type of the designated member.

¶4 A postfix expression followed by the -> operator and an identifier designates a member of a structure or union object. The value is that of the named member of the object to which the first expression points, and is an lvalue.⁹⁶⁾ If the first expression is a pointer to a qualified type, the result has the so-qualified version of the type of the designated member.

¶5 …

¶6 One special guarantee is made in order to simplify the use of unions: if a union contains several structures that share a common initial sequence (see below), and if the union object currently contains one of these structures, it is permitted to inspect the common initial part of any of them anywhere that a declaration of the completed type of the union is visible. Two structures share a common initial sequence if corresponding members have compatible types (and, for bit-fields, the same widths) for a sequence of one or more initial members.

---

⁹⁵⁾ If the member used to read the contents of a union object is not the same as the member last used to store a value in the object, the appropriate part of the object representation of the value is reinterpreted as an object representation in the new type as described in 6.2.6 (a process sometimes called ‘‘type punning’’). This might be a trap representation.

⁹⁶⁾ If &E is a valid pointer expression (where & is the ‘‘address-of’’ operator, which generates a pointer to its operand), the expression (&E)->MOS is the same as E.MOS.

标准中的斜体
第 6.2.6 节 类型的表示说(部分):

§6.2.6.1 General

¶6 When a value is stored in an object of structure or union type, including in a member object, the bytes of the object representation that correspond to any padding bytes take unspecified values.⁵¹⁾ The value of a structure or union object is never a trap representation, even though the value of a member of the structure or union object may be a trap representation.

¶7 When a value is stored in a member of an object of union type, the bytes of the object representation that do not correspond to that member but do correspond to other members take unspecified values.

---

⁵¹⁾ Thus, for example, structure assignment need not copy any padding bits.

我对您正在做的事情的解释是脚注 51 说“它可能不起作用”，因为您可能只分配了结构的一部分。充其量，你是在如履薄冰。但是，与此相反，您规定分配的结构(在 f.my_struct = d.my_struct; 分配中)是最大的成员。它不会出错的可能性相当高，但是如果两个结构中的填充字节(在 union 的事件成员和 union 的最大成员中)位于不同的位置，那么事情可能会出错并且如果您向编译器编写者报告了问题，编译器编写者只会对您说“不要违反标准”。
所以，就我是一名语言律师而言，这位语言律师的回答是“不能保证”。在实践中，您不太可能遇到问题，但可能性是存在的，而且您不会对任何人卷土重来。
为了使您的代码安全，只需使用 f = d;与 union 任务。

说明性示例
假设机器需要 double在 8 字节边界和 sizeof(double) == 8 上对齐, 那 int必须在 4 字节边界上对齐，并且 sizeof(int) == 4 ，以及 short必须在 2 字节边界上对齐，并且 sizeof(short) == 2 )。这是一组合理甚至常见的尺寸和对齐要求。
此外，假设您有问题中结构的双结构 union 变体:

struct Type_A { char x; double y; };
struct Type_B { int a; short b; short c; };
enum whichmember { TYPE_A, TYPE_B };

struct foo
{
    enum whichmember w;
    union
    {
        struct Type_A s1;
        struct Type_B s2;
    };
};

现在，在指定的大小和对齐方式下， struct Type_A将占用 16 个字节，而 struct Type_B将占用 8 个字节，因此 union 也将使用 16 个字节。 union 的布局将是这样的:

+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| x | p...a...d...d...i...n...g |               y               |  s1
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
|       a       |   b   |   c   |   p...a...d...d...i...n...g   |  s2
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+

w元素也意味着 struct foo 中有 8 个字节在(匿名) union 之前，其中很可能是 w只占4个。 struct foo的大小因此在这台机器上是 24。不过，这与讨论并不特别相关。
现在假设我们有这样的代码:

struct foo d;
d.w = TYPE_B;
d.s2.a = 1234;
d.s2.b = 56;
d.s2.c = 78;

struct foo f;
f.s1 = d.s1;
f.w  = TYPE_B;

现在，根据脚注 51，结构分配 f.s1 = d.s1;不必复制填充位。我知道没有编译器会这样，但标准说编译器不需要复制填充位。这意味着 f.s1 的值可能:

+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| x | g...a...r...b...a...g...e |   r...u...b...b...i...s...h   |
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+

垃圾是因为这 7 个字节不需要被复制(脚注 51 说这是一个选项，即使它不太可能是任何当前编译器执行的选项)。垃圾是因为 d的初始化永远不要在这些字节中设置任何值；该结构部分的内容未指定。
如果你现在继续尝试治疗 f作为 d 的副本，您可能会惊讶地发现 f.s2 的 8 个相关字节中只有 1 个字节实际上是初始化的。
我会再次强调:我知道没有编译器会这样做。但是这个问题被标记为“语言律师”，所以问题是“语言标准说明了什么”，这是我对标准引用部分的解释。