布局兼容类型的并集 | 布局兼容类型的并集

本文介绍了布局兼容类型的并集的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

看下面的代码:

struct A {
    short s;
    int i;
};
struct B {
    short s;
    int i;
};

union U {
    A a;
    B b;
};

int fn() {
    U u;
    u.a.i = 1;
    return u.b.i;
}

是否保证fn()返回1?

注意:这是对此的后续问题

Note: this is a follow-up question to this.

推荐答案

是的，这是定义的行为.首先让我们看看标准关于A和B的内容. [class.prop]/3 有

Yes, this is defined behavior. First lets see what the standard has to say about A and B. [class.prop]/3 has

没有非标准布局类(或此类数组)或引用的非静态数据成员，
没有虚拟函数，也没有虚拟基类，
对所有非静态数据成员具有相同的访问控制，
没有非标准布局的基类，
最多具有一个任何给定类型的基类子对象，
具有该类及其基类中所有的非静态数据成员和位字段，并且首先在同一类中声明它们，并且
[...](在这种情况下，这里什么也没说)

has no non-static data members of type non-standard-layout class (or array of such types) or reference,
has no virtual functions and no virtual base classes,
has the same access control for all non-static data members,
has no non-standard-layout base classes,
has at most one base class subobject of any given type,
has all non-static data members and bit-fields in the class and its base classes first declared in the same class, and
[...] (nothing said here has any bearing in this case)

所以A和B都是标准布局类型.如果我们查看 [class.mem]/23

So A and B are both standard layout types. If we look at [class.mem]/23

和 [class.mem]/22

和 [class.mem]/25

struct T1 { int a, b; };
struct T2 { int c; double d; };
union U { T1 t1; T2 t2; };
int f() {
  U u = { { 1, 2 } };   // active member is t1
  return u.t2.c;        // OK, as if u.t1.a were nominated
}

-示例示例"] [注:通过非易失性类型的glvalue读取易失性对象具有未定义的行为([dcl.type.cv]). —尾注]

— end example ] [ Note: Reading a volatile object through a glvalue of non-volatile type has undefined behavior ([dcl.type.cv]). — end note ]

那么我们就有了这样的类，它们具有相同的公共初始序列，布局相同，并且访问非活动类型的相同成员就像访问活动类型的成员一样.

Then we have that the classes have the same common initial sequence, are laid out the same, and accessing the same member of the non-active type is treated as if accessing that member of the active type.

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