C和c++中联合的目的

我以前很轻松地使用过工会;今天当我读到这篇文章并知道这个代码时，我很震惊

union ARGB
{
    uint32_t colour;

    struct componentsTag
    {
        uint8_t b;
        uint8_t g;
        uint8_t r;
        uint8_t a;
    } components;

} pixel;

pixel.colour = 0xff040201;  // ARGB::colour is the active member from now on

// somewhere down the line, without any edit to pixel

if(pixel.components.a)      // accessing the non-active member ARGB::components

实际上是未定义的行为，即从工会成员中读取除最近写的人以外的内容会导致未定义的行为。如果这不是联合的预期用途，那么什么才是?谁能详细解释一下吗?

更新:

我想事后澄清一些事情。

The answer to the question isn't the same for C and C++; my ignorant younger self tagged it as both C and C++. After scouring through C++11's standard I couldn't conclusively say that it calls out accessing/inspecting a non-active union member is undefined/unspecified/implementation-defined. All I could find was §9.5/1: If a standard-layout union contains several standard-layout structs that share a common initial sequence, and if an object of this standard-layout union type contains one of the standard-layout structs, it is permitted to inspect the common initial sequence of any of standard-layout struct members. §9.2/19: Two standard-layout structs share a common initial sequence if corresponding members have layout-compatible types and either neither member is a bit-field or both are bit-fields with the same width for a sequence of one or more initial members. While in C, (C99 TC3 - DR 283 onwards) it's legal to do so (thanks to Pascal Cuoq for bringing this up). However, attempting to do it can still lead to undefined behavior, if the value read happens to be invalid (so called "trap representation") for the type it is read through. Otherwise, the value read is implementation defined. C89/90 called this out under unspecified behavior (Annex J) and K&R's book says it's implementation defined. Quote from K&R: This is the purpose of a union - a single variable that can legitimately hold any of one of several types. [...] so long as the usage is consistent: the type retrieved must be the type most recently stored. It is the programmer's responsibility to keep track of which type is currently stored in a union; the results are implementation-dependent if something is stored as one type and extracted as another. Extract from Stroustrup's TC++PL (emphasis mine) Use of unions can be essential for compatness of data [...] sometimes misused for "type conversion".

最重要的是，这个问题(它的标题从我的提问开始就没有改变)是为了理解联合的目的而提出的，而不是关于标准允许什么。例如，使用继承来实现代码重用当然是c++标准允许的，但这并不是将继承引入c++语言特性的目的或初衷。这就是为什么安德烈的回答仍然被人们所接受的原因。

当前回答

再举一个联合实际使用的例子，CORBA框架使用带标签的联合方法序列化对象。所有用户定义的类都是一个(巨大的)联合的成员，整数标识符告诉解编码器如何解释该联合。

2010-02-22 19:42:18

其他回答

其他人提到了架构上的差异(小端到大端)。

我读到的问题是，由于变量的内存是共享的，那么写入一个变量，其他变量就会改变，根据它们的类型，值可能是没有意义的。

如。联盟{ 浮动f; int我; });

如果你从x.f读取数据，那么写入x.i是没有意义的——除非你想要查看浮点数的符号、指数或尾数分量。

我认为还有一个对齐的问题:如果一些变量必须字对齐，那么你可能得不到预期的结果。

如。联盟{ 字符c [4]; int我; });

假设，在某些机器上，一个char必须字对齐，那么c[0]和c[1]将与i共享存储空间，而不是c[2]和c[3]。

2010-02-22 14:25:14

从语言的角度来看，行为是未定义的。考虑到不同的平台在内存对齐和字节序方面可能有不同的约束。大端序机器中的代码与小端序机器中的代码将以不同的方式更新结构中的值。修复语言中的行为将要求所有实现使用相同的字节序(和内存对齐约束……)来限制使用。

如果你正在使用c++(你正在使用两个标签)，你真的关心可移植性，那么你可以只使用结构和提供一个setter，采用uint32_t和设置字段适当通过位掩码操作。在C语言中用函数也可以做到这一点。

Edit: I was expecting AProgrammer to write down an answer to vote and close this one. As some comments have pointed out, endianness is dealt in other parts of the standard by letting each implementation decide what to do, and alignment and padding can also be handled differently. Now, the strict aliasing rules that AProgrammer implicitly refers to are a important point here. The compiler is allowed to make assumptions on the modification (or lack of modification) of variables. In the case of the union, the compiler could reorder instructions and move the read of each color component over the write to the colour variable.

2010-02-22 11:28:27

@bobobobo代码是正确的，正如@Joshua指出的那样(遗憾的是，我不允许添加注释，所以在这里做，IMO不允许它放在第一位的坏决定):

https://en.cppreference.com/w/cpp/language/data_members#Standard_layout告诉我们这样做是可以的，至少从c++ 14开始

在具有非并集类类型T1的活动成员的标准布局联合中，允许读取另一个非并集类类型T2的联合成员的非静态数据成员m，前提是m是T1和T2的公共初始序列的一部分(除非通过非易失性glvalue读取易失性成员是未定义的)。

因为在当前的情况下T1和T2无论如何都提供了相同的类型。

2020-10-05 13:49:29

从技术上讲，它是未定义的，但实际上大多数(所有?)编译器都将其视为从一种类型使用reinterpret_cast到另一种类型，其结果是已定义的实现。我不会因为你现在的代码而失眠。

2010-02-22 11:49:28

你可以使用联合来创建像下面这样的结构体，它包含一个字段，告诉我们联合的哪个组件实际被使用:

struct VAROBJECT
{
    enum o_t { Int, Double, String } objectType;

    union
    {
        int intValue;
        double dblValue;
        char *strValue;
    } value;
} object;

2010-02-22 11:24:30

C和c++中联合的目的

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