How can I deal with @objc inference deprecation with #selector() in Swift 4?

As already mentioned in other answers, there is no way to avoid the @objc annotation for selectors.

But warning mentioned in the OP can be silenced by taking following steps:

1. Go to Build Settings
2. Search for keyword @objc
3. Set the value of Swift 3 @objc interface to Off

below is the screenshot that illustrates the above mentioned steps:

Hope this helps

As of, I think Swift 4.2, all you need to do is assign @IBAction to your method and avoid the @objc annotation.

let tap  =  UITapGestureRecognizer(target: self, action: #selector(self.cancel))

@IBAction func cancel()
{
    self.dismiss(animated: true, completion: nil)
}

The fix-it is correct – there's nothing about the selector you can change in order to make the method it refers to exposed to Objective-C.

The whole reason for this warning in the first place is the result of SE-0160. Prior to Swift 4, internal or higher Objective-C compatible members of NSObject inheriting classes were inferred to be @objc and therefore exposed to Objective-C, therefore allowing them to be called using selectors (as the Obj-C runtime is required in order to lookup the method implementation for a given selector).

However in Swift 4, this is no longer the case. Only very specific declarations are now inferred to be @objc, for example, overrides of @objc methods, implementations of @objc protocol requirements and declarations with attributes that imply @objc, such as @IBOutlet.

The motivation behind this, as detailed in the above linked proposal, is firstly to prevent method overloads in NSObject inheriting classes from colliding with each other due to having identical selectors. Secondly, it helps reduce the binary size by not having to generate thunks for members that don't need to be exposed to Obj-C, and thirdly improves the speed of dynamic linking.

If you want to expose a member to Obj-C, you need to mark it as @objc, for example:

class ViewController: UIViewController {

    @IBOutlet weak var button: UIButton!

    override func viewDidLoad() {
        super.viewDidLoad()
        button.addTarget(self, action: #selector(foo), for: .touchUpInside)
    }

    @objc func foo() {
       // ... 
    }
}

(the migrator should do this automatically for you with selectors when running with the "minimise inference" option selected)

To expose a group of members to Obj-C, you can use an @objc extension:

@objc extension ViewController {

    // both exposed to Obj-C
    func foo() {}
    func bar() {}
}

This will expose all the members defined in it to Obj-C, and give an error on any members that cannot be exposed to Obj-C (unless explicitly marked as @nonobjc).

If you have a class where you need all Obj-C compatible members to be exposed to Obj-C, you can mark the class as @objcMembers:

@objcMembers
class ViewController: UIViewController {
   // ...
}

Now, all members that can be inferred to be @objc will be. However, I wouldn't advise doing this unless you really need all members exposed to Obj-C, given the above mentioned downsides of having members unnecessarily exposed.

As Apple Official Documentation. you need to use @objc to call your Selector Method.

In Objective-C, a selector is a type that refers to the name of an Objective-C method. In Swift, Objective-C selectors are represented by the Selector structure, and can be constructed using the #selector expression. To create a selector for a method that can be called from Objective-C, pass the name of the method, such as #selector(MyViewController.tappedButton(sender:)). To construct a selector for a property’s Objective-C getter or setter method, pass the property name prefixed by the getter: or setter: label, such as #selector(getter: MyViewController.myButton).