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。
