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

Question

I asked a question here: Lifetime Extension of a initializer_list return involving the non-functional code:

const auto foo = [](const auto& a, const auto& b, const auto& c) { return {a, b, c}; };

I believed the lambda was trying to return an intializer_list (that's bad, don't do that.) But I got a comment:

It's not an initializer_list, it's an initializer list. Two different things.

I just thought that any time you did a curly-braced list you were creating an intializer_list. If that's not what's happening, what is a list in curly-braces?

Answer 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?