Allow the omission of the superclass

Currently the syntax is:

class MyClass extends mix(MySuperClass).with(MyMixin1, MyMixin2) {
}

How about the following?

class MyClass extends mixins(MyMixin1, MyMixin2, MySuperClass) {
}

Details:

• This better reflects the order of the inheritance chain
• The last parameter should be optional (you may want to create classes that only assemble mixins and have no superclass)
• The function could do a simple typecheck by enforcing that the mixins must not have a property prototype, while the superclass (the optional last parameter) must have this property.

[psayre23]

If this syntax is acceptable:

class MyClass extends mix().with(MyMixin1, MyMixin2) {

…then you could throw this check in for mix constructor:

if(typeof superclass !== "object" || superclass === null) {
    superclass = Object;
}

That would default to having Object as a superclass…which everything is, so that’s technically true.

[trusktr]

I was just thinking about this, as I thought “what if I don’t have a super class, and just want to extend from some mixins?”.

if (typeof superclass !== "object" || superclass === null) {
    superclass = Object;
}

That would default to having Object as a superclass…which everything is, so that’s technically true.

Object can be replaced at runtime, whereas class {} always extends from the internal [[Object]], so possibly it’s better to use an empty class anyway.

if (typeof superclass !== "object" || superclass === null) {
    superclass = class {};
}