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c++c++11initializationlanguage-lawyerc++-faq
Adrian McCarthy
Adrian McCarthy

Reputation: 48038

Meaning of default initialization changed in C++11?

C++2003 8.5/5 says:

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.

[Emphasis added.]

The C++2011 standard changed that last item to

— otherwise, no initialization is performed.

This seems like it would be a breaking change for some programs. Was this intentional?

Edit

Here's some code to motivate this question:

class Foo {
  public:
    Foo() : m_values() {}

    int m_values[3];
};

Before C++11, I thought the explicit mention of m_values in the default constructor would default-initialize that array. And since the elements of the array are scalar, I expected that to mean the values were all set to 0.

In C++11, it seems there's no longer a guarantee that this will happen. But maybe, as Mooing Duck pointed out in the comments, perhaps this is no longer a case of default initialization but some other form which preserves the expected behavior. Citations welcome.

Upvotes: 36

Views: 4655

Answers (3)

MWid
MWid

Reputation: 4589

Strictly speaking, the definition of default-initialize has changed from C++03 to C++11. But one has also to take into account that the situations when an object is _default-initialize_d changed:

§8.5p9 C++03 states:

If no initializer is specified for an object, and the object is of (possibly cv-qualified) non-POD class type (or array thereof), the object shall be default-initialized; if the object is of const-qualified type, the underlying class type shall have a user-declared default constructor. Otherwise, if no initializer is specified for a nonstatic object, the object and its subobjects, if any, have an indeterminate initial value; if the object or any of its subobjects are of const-qualified type, the program is ill-formed.

§8.5p11 C++11 states:

If no initializer is specified for an object, the object is default-initialized; if no initialization is performed, an object with automatic or dynamic storage duration has indeterminate value.

As @JamesKanze already pointed out, default-initialization is performed in C++03 when no initializer for an object of non-POD class type is specified. In C++11 an object (of arbitrary type) is default-initialized if no initializer is specified. Because of this change, the definition of default-initialize had also to be changed in order to be compatible with C++03.

Your example has nothing to do with default-initialization. It has been always the case that an object whose initializer is an empty set of parentheses is value-initialized.

Upvotes: 2

user1508519
user1508519

Reputation:

According to cppreference.com (because it uses friendlier language than the standard):

Default initialization is performed in three situations:

3) when a base class or a non-static data member is not mentioned in a constructor initializer list and that constructor is called.

Value initialization is performed in three situations:

3,7) when a non-static data member or a base class is initialized using a member initializer with an empty pair of parentheses or braces (since C++11)

Note that the C++11 part belongs with the or braces, not with the entire paragraph.

And:

To value-initialize an object of type T means:
— if T is an array type, then each element is value-initialized;
— otherwise, the object is zero-initialized

So in C++11 default-initialization does not zero-initialize members but value-initialization does.

Upvotes: 3

James Kanze
James Kanze

Reputation: 154047

The final effects are almost the same. In C++03, the use of default-initialize was restricted to non-POD class type, so the last point never applied. In C++11, the standard simplifies the wording by eliminating the condition with regards to where default-initialization was used, and changes the definition of default-initialization to cover all of the cases in a way to correspond what happened before.

Upvotes: 25

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