The question betrays some misunderstandings about what is going on here.
std::array is an array object. It is an object whose storage size is the storage for its array elements. std::initializer_list is a pointer to an array (one created by the compiler). You can heap-allocate std::array for example; you cannot heap-allocate std::initializer_list. Well, you can, but you’d just be heap allocating a pointer, which usually isn’t helpful. This:
auto *p = new std::initializer_list<int>{1, 2, 3, 4};
Is broken code, as it heap allocates a pointer to an array temporary created by the compiler. An array temporary that is immediately destroyed at the end of this statement. So you now have a pointer to an array that does not exist, so using *p or p[x] is undefined behavior.
Doing the same with std::array works fine, because array is an array.
Furthermore, the purpose of std::initializer_list is not to “define a list at compile time”. As the name suggests, the point of the type is to create a list for the purpose of initializing an object. This is why the ultimate source of its data cannot be provided by any means other than a specific piece of C++ grammar, one used specifically to initialize objects: a braced-init-list.
You can use std::initializer_list as a quick-and-dirty way of just creating an array of values for some purpose. But that’s not the point of the type. That’s why it doesn’t have an operator[]; because functions that can be initialized from a sequence of values don’t tend to need that particular operation. They usually only need to walk it from beginning to end.
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solved Why did STL made a(nother) distinction between a std::array and a std::initializer_list