Initializing a union with a non-trivial constructor
Solution 1:
Question 1: Default constructors do initialize POD members to 0 according to the C++ standard. See the quoted text below.
Question 2: If a constructor must be specified in a base class, then that class cannot be part of a union.
Finally, you can provide a constructor for your union:
union U
{
A a;
B b;
U() { memset( this, 0, sizeof( U ) ); }
};
For Q1:
From C++03, 12.1 Constructors, pg 190
The implicitly-defined default constructor performs the set of initializations of the class that would be performed by a user-written default constructor for that class with an empty mem-initializer-list (12.6.2) and an empty function body.
From C++03, 8.5 Initializers, pg 145
To default-initialize an object of type T means:
- if T is a non-POD class type (clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
- if T is an array type, each element is default-initialized;
- otherwise, the object is zero-initialized.
To zero-initialize an object of type T means:
- if T is a scalar type (3.9), the object is set to the value of 0 (zero) converted to T;
- if T is a non-union class type, each non static data member and each base-class subobject is zero-initialized;
- if T is a union type, the object’s first named data member is zero-initialized;
- if T is an array type, each element is zero-initialized;
- if T is a reference type, no initialization is performed.
For Q2:
From C++03, 12.1 Constructors, pg 190
A constructor is trivial if it is an implicitly-declared default constructor and if:
- its class has no virtual functions (10.3) and no virtual base classes (10.1), and
- all the direct base classes of its class have trivial constructors, and
- for all the nonstatic data members of its class that are of class type (or array thereof), each such class has a trivial constructor
From C++03, 9.5 Unions, pg 162
A union can have member functions (including constructors and destructors), but not virtual (10.3) functions. A union shall not have base classes. A union shall not be used as a base class.An object of a class with a non-trivial constructor (12.1), a non-trivial copy constructor (12.8), a non-trivial destructor (12.4), or a non-trivial copy assignment operator (13.5.3, 12.8) cannot be a member of a union, nor can an array of such objects
Solution 2:
Things changed for the better in C++11.
You can now legally do this, as described by Stroustrup himself (I reached that link from the Wikipedia article on C++11).
The example on Wikipedia is as follows:
#include <new> // Required for placement 'new'.
struct Point {
Point() {}
Point(int x, int y): x_(x), y_(y) {}
int x_, y_;
};
union U {
int z;
double w;
Point p; // Illegal in C++03; legal in C++11.
U() {new(&p) Point();} // Due to the Point member, a constructor
// definition is now *required*.
};
Stroustrup goes into a little more detail.
Solution 3:
As mentioned in Greg Rogers' comment to unwesen's post, you can give your union a constructor (and destructor if you wish):
struct foo
{
int a;
int b;
};
union bar
{
bar() { memset(this, 0, sizeof(*this)); }
int a;
foo f;
};
Solution 4:
AFAIK union members may not have constructors or destructors.
Question 1: no, there's no such guarantee. Any POD-member not in the constructor's initialization list gets default-initialized, but that's with a constructor you define, and has an initializer list. If you don't define a constructor, or you define a constructor without an initializer list and empty body, POD-members will not be initialized.
Non-POD members will always be constructed via their default constructor, which if synthesized, again would not initialize POD-members. Given that union members may not have constructors, you'd pretty much be guaranteed that POD-members of structs in a union will not be initialized.
Question 2: you can always initialize structures/unions like so:
struct foo
{
int a;
int b;
};
union bar
{
int a;
foo f;
};
bar b = { 0 };