Does copy list initialization invoke copy ctor conceptually?

Question

Before C++11, we can do copy initialization by writing something like A a = 1; which is more or less equivalent to A a = A(1);. That is, a temporar

Answer 1

The standard describes it pretty well; [dcl.init.list]/3:

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

• [...]
• Otherwise, if T is a class type, constructors are considered. The applicable constructors are enumerated and the best one is chosen through overload resolution (13.3, 13.3.1.7). If a narrowing conversion (see below) is required to convert any of the arguments, the program is ill-formed.

[over.match.list] (emphasis mine):

When objects of non-aggregate class type T are list-initialized (8.5.4), overload resolution selects the constructor in two phases:

• Initially, the candidate functions are the initializer-list constructors (8.5.4) of the class T and the argument list consists of the initializer list as a single argument.

• If no viable initializer-list constructor is found, overload resolution is performed again, where the candidate functions are all the constructors of the class T and the argument list consists of the elements of the initializer list.

If the initializer list has no elements and T has a default constructor, the first phase is omitted.
In copy-list-initialization, if an explicit constructor is chosen, the initialization is ill-formed.

Hence, if no initializer-list constructor is found (as in your case), the elements of the initializer list constitute the arguments for the constructor call.
In fact, the only difference of direct-list-initialization and copy-list-initialization is covered by the last, bolded sentence.

This is one of the advantages of list-initialization: It doesn't necessitate the presence of a special member function that is not gonna be used anyway.