Why must <initializer_list> be included for using auto?

Question

There has already been a similar question on SO, but I want to stress another aspect of braced-init-lists. Consider the following:

auto x = {1}; //(1)

This is ill-formed (8.5.4/2) unless the header <initializer_list> is included. But why? The standard says, that the template std::initializer_list is not predefined. Does this mean, that declaration (1) introduces a new type? In all other situations, where auto may be used such as

auto y = expr;

where expr is an expression, the type auto deduces already exists. On the other hand, from a logical point of view, the compiler must assign an implicite type to the construct {1}, for which std::initializer_list is then another name. But in declaration (1) we do not want to name this type. So why must this header be included. There is a similar situation with nullptr. Its type implicitely exists, but to name it explicitely you have to include <cstddef>.

Answer 1

That's not the same. The rules for std::nullptr_t and std::initializer_list are actually different.

std::nullptr_t is just a typedef for a built-in type. Its definition is

namespace std {
  using nullptr_t = decltype(nullptr);
}

The type exists whether you include the header or not.

std::initializer_list is a class template, not a predefined type. It really doesn't exist unless you include the header that defines it. In particular, the initializer list { 1 } does not have type std::initializer_list<int>; it has no type at all, because it is not an expression. (Initializer lists are special syntactic constructs and cannot appear everywhere an expression can.)

std::initializer_list is just slightly special. For one, there are special rules for how to initialize a std::initializer_list from the initializer list syntax (allocate an array and have the object refer to it). However, this requires std::initializer_list to be defined in the first place.

The second special case is auto type deduction. There's a special rule here too. But again, this doesn't mean that the compiler will automatically define the type; it just means that it will recognize it.