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

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

其他回答

正如你所说，这是严格未定义的行为，尽管它将“工作”在许多平台上。使用联合的真正原因是为了创建不同的记录。

union A {
   int i;
   double d;
};

A a[10];    // records in "a" can be either ints or doubles 
a[0].i = 42;
a[1].d = 1.23;

当然，您还需要某种鉴别器来说明这个变体实际上包含了什么。注意，在c++中，联合的用处不大，因为它们只能包含POD类型——实际上是那些没有构造函数和析构函数的类型。

2010-02-22 11:22:20

在c++中，Boost Variant实现了一个安全的联合版本，旨在尽可能地防止未定义的行为。

它的性能与enum + union结构相同(也分配了堆栈等)，但它使用类型的模板列表而不是enum:)

2010-02-22 13:09:57

行为可能没有定义，但这只是意味着没有一个“标准”。所有优秀的编译器都提供#pragmas来控制打包和对齐，但可能有不同的默认值。默认值也会根据所使用的优化设置而改变。

此外，工会不仅仅是为了节省空间。它们可以帮助现代编译器使用类型双关语。如果你reinterpret_cast<>所有的东西，编译器就不能假设你正在做什么。它可能不得不放弃它所知道的类型并重新开始(强制写回内存，与CPU时钟速度相比，这是非常低效的)。

2012-01-18 11:21:31

正如其他人提到的，联合与枚举结合并包装成结构体可用于实现带标签的联合。一个实际用途是实现Rust的Result<T, E>，它最初是使用纯枚举实现的(Rust可以在枚举变量中保存额外的数据)。下面是一个c++的例子:

template <typename T, typename E> struct Result {
    public:
    enum class Success : uint8_t { Ok, Err };
    Result(T val) {
        m_success = Success::Ok;
        m_value.ok = val;
    }
    Result(E val) {
        m_success = Success::Err;
        m_value.err = val;
    }
    inline bool operator==(const Result& other) {
        return other.m_success == this->m_success;
    }
    inline bool operator!=(const Result& other) {
        return other.m_success != this->m_success;
    }
    inline T expect(const char* errorMsg) {
        if (m_success == Success::Err) throw errorMsg;
        else return m_value.ok;
    }
    inline bool is_ok() {
        return m_success == Success::Ok;
    }
    inline bool is_err() {
        return m_success == Success::Err;
    }
    inline const T* ok() {
        if (is_ok()) return m_value.ok;
        else return nullptr;
    }
    inline const T* err() {
        if (is_err()) return m_value.err;
        else return nullptr;
    }

    // Other methods from https://doc.rust-lang.org/std/result/enum.Result.html

    private:
    Success m_success;
    union _val_t { T ok; E err; } m_value;
}

2019-05-27 16:16:10

尽管这是严格未定义的行为，但实际上它适用于几乎任何编译器。它是一种被广泛使用的范例，任何有自尊的编译器都需要在这种情况下做“正确的事情”。它当然比类型双关语更受欢迎，在某些编译器中，类型双关语很可能会生成坏代码。

2010-02-22 11:22:40

C和c++中联合的目的

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