c++ - 使用 union 的一个字段的地址访问另一个字段是否合法？

Question

考虑以下代码:

union U
{
    int a;
    float b;
};

int main()
{
    U u;
    int *p = &u.a;
    *(float *)p = 1.0f; // <-- this line
}

我们都知道 union 字段的地址通常是相同的，但我不确定这样做是否是明确定义的行为。

因此，问题是:像上面的代码一样强制转换和取消引用指向 union 字段的指针是否合法且定义明确的行为？

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附言我知道它更像是 C 语言而不是 C++，但我想了解它在 C++ 中是否合法，而不是 C。

Answer 1

联盟的所有成员必须位于同一地址，这是标准所保证的。您正在做的确实是明确定义的行为，但应该注意的是，您不能使用相同的方法从 union 的非事件成员中读取数据。

• Accessing inactive union member - undefined behavior?

_{Note: Do not use c-style casts, prefer reinterpret_cast in this case.}

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只要您所做的只是写入到union 的其他数据成员，行为就是明确定义的；但如前所述，这种变化被认为是 union 的 active 成员；这意味着您以后只能读取刚刚写入的内容。

union U {
    int a;
    float b;
};

int main () {
    U u;
    int *p = &u.a;
    reinterpret_cast<float*> (p) = 1.0f; // ok, well-defined
}

_{Note: There is an exception to the above rule when it comes to layout-compatible types.}

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问题可以改写为以下片段，这在语义上等同于“问题”的简化版本。

#include <type_traits>
#include <algorithm>
#include <cassert>

int main () {
  using union_storage_t = std::aligned_storage<
    std::max ( sizeof(int),   sizeof(float)),
    std::max (alignof(int),  alignof(float))
  >::type;

  union_storage_t u;

  int   * p1 = reinterpret_cast<  int*> (&u);
  float * p2 = reinterpret_cast<float*> (p1);
  float * p3 = reinterpret_cast<float*> (&u);

  assert (p2 == p3); // will never fire
}

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标准 ( n3797 ) 说了什么？

9.5/1    <b>Unions</b>    [class.union]

In a union, at most one of the non-static data members can be active at any time, that is, the value of at most one of the non-static dat amembers ca nbe stored in a union at any time. [...] The size of a union is sufficient to contain the largest of its non-static data members. Each non-static data member is allocated as if it were the sole member of a struct. All non-static data members of a union object have the same address.

_{Note: The wording in C++11 (n3337) was underspecified, even though the intent has always been that of C++14.}