Union of layout-compatible types

问题

Look at this code:

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;
}

Is it guaranteed that fn() returns 1?

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

回答1:

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

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)

So A and B are both standard layout types. If we look at [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]).

and [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.

and [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 ]

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.

来源：https://stackoverflow.com/questions/53051088/union-of-layout-compatible-types

标签

c++

language-lawyer

c++17

unions

standard-layout