Exactly, needs better documentation.

Cheddar Bob

@@nijucow i ALWAYS hear: " My name is 'Ocherno' "

I always hear it: "My name is the Cyano"

I was going to write same thing. :)

if I'm not mistaken reference types also must be initialized in the initializer list

What about static .?

@@venkateswarans1012 Well, those will be initialized by default I think, because you can access them without creating an object

Basically all members which should be initialized at time of declaration should be written in initializer list like const , references etc.

confirmed

Greatgreat video though, I will forever use this from now on!

Basically, all data-members which should be initialized at time of declaration, such as const , references, etc., must be included in the constructor's initializer list.

Basically, all data-members which should be initialized at time of declaration, such as const , references, etc., must be included in the constructor's initializer list.

You just keep the order the same as your members to not confuse yourself. Try something like Entity:m_x(5),m_y(m_x+1) and flip the order.

i mean, he explicitly states that it could mess up depending on the compiler. and feels like he's mostly encouraging this for good practice.

Good summary

Dot is accessing members of an object eg Dog d; dog.age = 10; Arrow is for dereferencing a pointer, aka accessing the variable the pointer is pointing to. Dog dog1; Dog* pDog = &dog; pDog->bark();//Calls bark on the "dog1" object Colon is for accessing classes and functions inside of a namespace, or static variables and functions in classes, or for defining member functions and defining static class variables.

Hopson wow, quick and brief. Thanks :)

@@Hopsonn Little nitpicking: Arrow does not dereference the pointer and access the variable it points to, rather it does that *then* accesses a *member* (or method) of that variable it points to. So technically two operation in one, thus the variable should be of a type with members (or methods), eg. class, struct. So to be consistent with your first example: Dog is a class or struct instance: Dog d; d.age = 10; d2 is a pointer to a Dog instance: Dog *d2 = &d; d2->age = 11; // this arrow op. is just a nicer, shorter form of this: (*d2).age = 11; These last two lines does exactly the same. So a simple dereferencing of d2 is useful to get the Dog instance it points to, without accessing it's member. It is *d2, for example Dog d3 = Dog(*d2); Executing all the above, you'll end up with 2 distinct Dog instances, d and d3 d2 is just a pointer to d d.age is overwritten to 11 trough the pointer. d3 is copied from where d2 points to, so from d, thus d3.age is also 11.