Ambiguity between default-initialization and value-initialization

Question

I found many articles explaining the difference between "default-initialization and value-initialization" but in fact I didn't understand clearly.

Here's an example:

class A{
   public:
      int x;
};


int main(){
    A a;// default initialization so x has undefined value.
    A b = A(); // value initialization so x is a scalar thus it is value initialized to 0

}

Above it is OK as I guess but here:

int value = 4; // is this considered a value-initialization?

Please help me understand the major differences between the two forms of initializations.

Holdover

Looks like Copy Initialization Here's another link that (roughly) covers all of the different types of initialization you could run across – Kneehigh

A a; does not have default initialisation. I thought A a1{}; default initialises, and A a2{1}; value initialises, and A a3{a2}; copy initialises, although the above link states that A a1{}; is value initialised, so perhaps I'm just wrong. – Trainman

@Trainman Yes you're just wrong. A a1{}; is value initialization, and A a2{1}; and A a3{a2}; are direct initialization. – Guffaw

Answer 1

A a; is default initialization, as the effect the default constructor of A is used for initialization. Since the implicitly-generated default constructor of A does nothing, a.x has indeterminate value.

A() is value initialization,

if T is a class type with a default constructor that is neither user-provided nor deleted (that is, it may be a class with an implicitly-defined or defaulted default constructor), the object is zero-initialized and then it is default-initialized if it has a non-trivial default constructor;

Note the difference with default initialization, A has an implicitly-defined default constructor, and the object is zero-initialized; so the data memeber x of the temporary object A() will be initialized to 0.

A b = A(); is copy initialization, in concept b is initialized from the temporary object A(), so b.x will be initialized to 0 too. Note that because of copy elision, since C++17 b is guaranteed to be value-initialized directly; the copy/move construction is omitted.

int value = 4; is copy initialization too. value will be initialized to 4.

Siret

Do you mean if class A has a user-defined constructor then writing A a; is not default-initialization? – Holdover

@RamiYen No, A a; is still default initialization. And A() is still value initialization. But the effect of value initialization will change to be same as default initialization. Both the effect will be initialize the object by the user-defined default constructor, zero initialization doesn't participate in again. – Siret