Array Initialization Initializes all Elements


I was talking about C++ with a Java programmer the other day, as he had to work on a bit of C++ code. I discovered that it isn’t necessarily obvious that all the elements in an array are constructed when the array is initialized. That is however always the case.

For arrays of built-in types, the values aren’t actually set to a specific default (except for non-local and static arrays), and will end up having arbitrary values.

For arrays of a user-defined type, the default constructor is used for all the array elements. If you don’t want this behaviour, you need to use an array of pointers. All the elements will still be initialized, but the elements being initialized are now the pointers, not the actual objects. It is worth mentioning that pointers count as a built in type, and elements of non-local and static arrays will have undefined values. In particular, they are not initialized to point to NULL.

The reason why all the elements are initialized is that there is no such thing as a “null-object”. You can have a null-pointer not pointing to anything, but you cannot have an actual object that isn’t really an object, so to speak. This of course goes for Java as well, but here, an array of a user-defined type is actually an array of pointers, even though you don’t see any *s in the code. (Java uses pointer semantics for user-defined types and value semantics for built-in types.)

Here is a summary using actual C++ code:

int ints[10]; //Non-local array, all elements ==0
int func() {
    int ints[10]; //Local array, no default value is set, but all elements are fully usable ints.
    static int sints[10]; //Static local array, all elements are initialized to 0
    Foo foos[10]; //User-defined type, all elements are constructed using the default constructor Foo::Foo();
    Foo *foos[10]; //Local array, no default value is set. The pointers point all over the place.
}

If your class has no public default constructor, it is impossible to create an array of such objects:

class NoDefault {
private:
    NoDefault(); //Make default constructor private, and hence unusable
};
NoDefault nodefaults[10]; //Impossible!
NoDefault *pnodefaults[10]; //This is fine, no objects are constructed, only pointers.

Note also that you cannot have an array of references. The most obvious reason is that a reference must always refer to an object. There is no such thing as a “null-reference”. One could however imagine getting around this using an initializer list, like this:

    Foo f1, f2;
    Foo& foors[2] = {f1, f2};

Now we don’t try to initialize references that don’t refer to anything. This is still not allowed though. One reason is that you cannot point to a reference, so accessing elements in the array in a normal sense like foors[1] wouldn’t make sense. In C++, you can only have arrays of objects, and references are not objects. (Pointers are objects though.)

In summary:

  • Elements of arrays of a user-defined type are initialized using the default constructor.
  • Elements of local non-static arrays of a built-in type are not explicitly initialized, and will have arbitrary values.
  • Elements of non-local and static arrays of a built-in type are initialized to their default value (0).
  • You cannot have an array of a user-defined type without a public default constructor.
  • You can only have arrays of objects and pointers, not references.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: