Initializing a union with a non-trivial constructor

I have a structure which I create a custom constructor to initialize the members to 0's. I've seen in older compilers that when in release mode, without doing a memset to 0, the values are not initialized.

I now want to use this structure in a union, but get errors because it has a non-trivial constructor.

So, question 1. Does the default compiler implemented constructor guarantee that all members of a structure will be null initialized? The non-trivial constructor just does a memset of all the members to '0' to ensure a clean structure.

Question 2: If a constructor must be specified on the base structure, how can a union be implemented to contain that element and ensure a 0 initialized base element?

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

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.

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 };

As mentioned in Greg Rogers' comment to https://stackoverflow.com/questions/321351/initializing-a-union-with-a-non-trivial-constructor#321466">unwesen</a>'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;
};

Can you do something like this?

class Outer
{
public:
	Outer()
	{
		memset(&inner_, 0, sizeof(inner_));
	}
private:
	union Inner
	{
		int qty_;
		double price_;
	} inner_;
};

...or maybe something like this?

union MyUnion
{
	int qty_;
	double price_;
};

void someFunction()
{
	MyUnion u = {0};
}

You'll have to wait for C++0x to be supported by compilers to get this. Until then, sorry.