Initializing a union with a non-trivial constructor

问题

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?

回答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

回答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.

回答3:

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 };

回答4:

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;
};

回答5:

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};
}

回答6:

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

来源：https://stackoverflow.com/questions/321351/initializing-a-union-with-a-non-trivial-constructor

标签

c++

constructor

multiplatform