What Is a Curly-Brace Enclosed List If Not an intializer_list?

Solution 1:

It is an braced-init-list. A braced-init-list existed before std::initializer_list and is used to initialize aggregates.

int arr[] = {1,2,3,4,5};

The above used a braced-init-list to initialize the array, no std::initializer_list is created. On the other hand when you do

std::vector<int> foo = {1,2,3,4,5};

foo is not an aggregate so the braced-init-list is used to create a std::initializer_list which is in turned passed to the constructor of foo that accepts a std::initializer_list.

A thing to note about a braced-init-list is that is has no type so special rules were developed for use with it and auto. It has the following behavior (since the adoption of N3922)

auto x1 = { 1, 2 }; // decltype(x1) is std::initializer_list<int>
auto x2 = { 1, 2.0 }; // error: cannot deduce element type
auto x3{ 1, 2 }; // error: not a single element
auto x4 = { 3 }; // decltype(x4) is std::initializer_list<int>
auto x5{ 3 }; // decltype(x5) is int

And you can get more information on history of this behavior and why it was changed at: Why does auto x{3} deduce an initializer_list?

Solution 2:

There are three distinct, but related concepts here:

1. braced-init-list: The grammatical rule associated with curly-brace-enclosed lists in certain contexts.

2. Initializer list: The name for the braced-init-list initializer used in list-initialization.

3. std::initializer_list: A class wrapping a temporary array which is created in some contexts involving braced-init-lists.

Some examples:

//a braced-init-list and initializer list, 
//but doesn't create a std::initializer_list
int a {4}; 

//a braced-init-list and initializer list,
//creates a std::initializer_list
std::vector b {1, 2, 3};

//a braced-init-list and initializer list,
//does not create a std::initializer_list (aggregate initialization)
int c[] = {1, 2, 3};

//d is a std::initializer_list created from an initializer list
std::initializer_list d {1, 2, 3};

//e is std::initializer_list<int>
auto e = { 4 };

//f used to be a std::initializer_list<int>, but is now int after N3922
auto f { 4 };

You might want to read N3922, which changed some of the rules involving auto and std::initializer_list.

Solution 3:

I just thought that any time you did a curly-braced list you were creating an intializer_list.

That's not correct.

If that's not what's happening, what is a list in curly-braces?

struct Foo {int a; int b;};
Foo f = {10, 20};

The {10, 20} part is not an initializer_list. It's just a syntactic form to use a list of objects to create another object.

int a[] = {10, 20, 30};

Once again, it's a syntactic form to create an array.

The name for the syntactic form is braced-init-list.