T v{} initialization

Question

I'm reading the C++11 standard, but can't figure out whether

T x{};

is value-initialized or default initialized (automatic storage). It does say pretty clearly that:

10 An object whose initializer is an empty set of parentheses, i.e., (), shall be value-initialized.

And that

11 If no initializer is specified for an object, the object is default-initialized;

But all I can find about T x{}; is that:

The initialization that occurs in the forms T x(a); T x{a}; as well as in new expressions (5.3.4), static_cast expressions (5.2.9), functional notation type conversions (5.2.3), and base and member initializers (12.6.2) is called direct-initialization.

and

If the initializer is a (non-parenthesized) braced-init-list, the object or reference is list-initialized (8.5.4).

I'm new to diving into the level of reading the standards. Can someone point me in the right direction?

Answer 1

This is indeed covered by your quote:

If the initializer is a (non-parenthesized) braced-init-list, the object or reference is list-initialized (8.5.4).

Skipping down to 8.5.4 List-initialization. Here I have paraphrased/omitted some points that don't pertain to the case of T x{}:

List-initialization of an object or reference of type T is defined as follows:

• If T is an aggregate, aggregate initialization is performed (8.5.1).
• Otherwise, if the initializer list has no elements and T is a class type with a default constructor, the object is value-initialized
• Otherwise, if T is a specialization of std::initializer_list<E> [...]
• Otherwise, [if the list is not empty and it matches a constructor]
• Otherwise, [if the list has a single element]
• Otherwise, [if T is a reference type]
• Otherwise, if the initializer list has no elements, the object is value-initialized.
• Otherwise, the program is ill-formed.

The first point, aggregate initialization was in C++03 as well; in that case T x{}; is the same as T x = {};.

For the second point "T is a class type with a default constructor", it is value-initialized which means calling the default constructor.

If T is a primitive type then the second-to-last point applies and it is value-initialized again.

Going back to the aggregate initialization case, in 8.5.1/7 there is:

If there are fewer initializer-clauses in the list than there are members in the aggregate, then each member not explicitly initialized shall be initialized from its brace-or-equal-initializer or, if there is no brace-or-equal-initializer, from an empty initializer list (8.5.4).

The brace-or-equal-initializer refers to an initializer provided inline in the class definition. If that isn't present then it is initialized as if the member had been initialized with {} (so, this logic is recursively applied for each aggregate member).

For example,

struct T
{
     int a;
};

then T x {}; leads to a being initialized as if it were int a{}; , which is value-initialization since int is a primitive type.