c++ - 布局兼容类型的 union

Question

看这段代码:

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 this 的后续问题.

Answer 1

是的，这是定义的行为。首先让我们看看标准对 A 和 B 有什么看法。 [class.prop]/3有

A class S is a standard-layout class if it:

• 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。如果我们看[class.mem]/23

Two standard-layout struct types are layout-compatible classes if their common initial sequence comprises all members and bit-fields of both classes ([basic.types]).

和[class.mem]/22

The common initial sequence of two standard-layout struct types is the longest sequence of non-static data members and bit-fields in declaration order, starting with the first such entity in each of the structs, such that corresponding entities have layout-compatible types, either both entities are declared with the no_­unique_­address attribute ([dcl.attr.nouniqueaddr]) or neither is, and either both entities are bit-fields with the same width or neither is a bit-field.

和[class.mem]/25

In a standard-layout union with an active member of struct type T1, it is permitted to read a non-static data member m of another union member of struct type T2 provided m is part of the common initial sequence of T1 and T2; the behavior is as if the corresponding member of T1 were nominated. [ Example:

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
}

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

然后我们得到类具有相同的公共(public)初始序列，布局相同，并且访问非事件类型的相同成员被视为访问事件类型的成员。