'#selector' refers to a method that is not exposed to Objective-C

In my case the function of the selector was private. Once I removed the private the error was gone. Same goes for fileprivate.

In Swift 4
You will need to add @objc to the function declaration. Until swift 4 this was implicitly inferred.


You need to use the @objc attribute on didTapCommentButton(_:) to use it with #selector.

You say you did that but you got another error. My guess is that the new error is that Post is not a type that is compatible with Objective-C. You can only expose a method to Objective-C if all of its argument types, and its return type, are compatible with Objective-C.

You could fix that by making Post a subclass of NSObject, but that's not going to matter, because the argument to didTapCommentButton(_:) will not be a Post anyway. The argument to an action function is the sender of the action, and that sender will be commentButton, which is presumably a UIButton. You should declare didTapCommentButton like this:

@objc func didTapCommentButton(sender: UIButton) {
    // ...
}

You'll then face the problem of getting the Post corresponding to the tapped button. There are multiple ways to get it. Here's one.

I gather (since your code says cell.commentButton) that you're setting up a table view (or a collection view). And since your cell has a non-standard property named commentButton, I assume it's a custom UITableViewCell subclass. So let's assume your cell is a PostCell declared like this:

class PostCell: UITableViewCell {
    @IBOutlet var commentButton: UIButton?
    var post: Post?

    // other stuff...
}

Then you can walk up the view hierarchy from the button to find the PostCell, and get the post from it:

@objc func didTapCommentButton(sender: UIButton) {
    var ancestor = sender.superview
    while ancestor != nil && !(ancestor! is PostCell) {
        ancestor = view.superview
    }
    guard let cell = ancestor as? PostCell,
        post = cell.post
        else { return }

    // Do something with post here
}

Tags:

Xcode

Swift