Meaning of default initialization changed in C++11?

Question

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.

Answer 1

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.