swift: How can I override a public method in superclass to be a private method in subclass

Question

I have a super class

class Father {
    public func doSomething() {
    } 
}

I want this child class to be

class Child: Father {
    private override func doSomething() {
    }
}

but Xcode says that

Overriding instance method must be as accessible as the declaration it overrides

so, how can I override a public method in superclass be a private method in subclass

THANK YOU

Answer 1

You can't because that would violate the Liskov Substitution Principle.

Essentially, any code that can operate on an instance of a superclass must also be able to operate on an instance of your subclass.

So, if some other class has a method

class Unrelated {
    func operateOnAThing(_ someThing:Father) {
        someThing.doSomething()
    }
}

then it still has to work when you do the following:

let aChild = Child()

unrelatedInstance.operateOnAThing(aChild)

If the doSomething method had more restrictive access in the Child class then you would get a runtime error. To prevent this, you cannot make access more restrictive in a subclass.

Answer 2

You can achieve that by marking the public method as unavailable using the @available attribute. Then, call the method using super in the private method.

Example:

class Father {
    public func doSomething() {
        print("doSomething privatly")
    }
}

class Child: Father {
    @available (*, unavailable)
    override func doSomething() {
    }

    fileprivate func internal_doSomething() {
        super.doSomething()
    }
}

Child().internal_doSomething()
Child().doSomething() //'doSomething()' has been explicitly marked unavailable

(Tested using Swift 4.0!)