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How to subclass UITableViewController in Swift

I want to subclass UITableViewController and be able to instantiate it by calling a default initializer with no arguments.

class TestViewController: UITableViewController {
    convenience init() {
        self.init(style: UITableViewStyle.Plain)
    }
}

As of the Xcode 6 Beta 5, the example above no longer works.

Overriding declaration requires an 'override' keyword
Invalid redeclaration of 'init()'
like image 323
Nick Snyder Avatar asked Aug 05 '14 13:08

Nick Snyder


3 Answers

NOTE This bug is fixed in iOS 9, so the entire matter will be moot at that point. The discussion below applies only to the particular system and version of Swift to which it is explicitly geared.


This is clearly a bug, but there's also a very easy solution. I'll explain the problem and then give the solution. Please note that I'm writing this for Xcode 6.3.2 and Swift 1.2; Apple has been all over the map on this since the day Swift first came out, so other versions will behave differently.

The Ground of Being

You are going to instantiate UITableViewController by hand (that is, by calling its initializer in code). And you want to subclass UITableViewController because you have instance properties you want to give it.

The Problem

So, you start out with an instance property:

class MyTableViewController: UITableViewController {
    let greeting : String
}

This has no default value, so you have to write an initializer:

class MyTableViewController: UITableViewController {
    let greeting : String
    init(greeting:String) {
        self.greeting = greeting
    }
}

But that's not a legal initializer - you have to call super. Let's say your call to super is to call init(style:).

class MyTableViewController: UITableViewController {
    let greeting : String
    init(greeting:String) {
        self.greeting = greeting
        super.init(style: .Plain)
    }
}

But you still can't compile, because you have a requirement to implement init(coder:). So you do:

class MyTableViewController: UITableViewController {
    let greeting : String
    required init(coder aDecoder: NSCoder) {
        fatalError("init(coder:) has not been implemented")
    }
    init(greeting:String) {
        self.greeting = greeting
        super.init(style: .Plain)
    }
}

Your code now compiles! You now happily (you think) instantiate this table view controller subclass by calling the initializer you wrote:

let tvc = MyTableViewController(greeting:"Hello there")

Everything looks merry and rosy until you run the app, at which point you crash with this message:

fatal error: use of unimplemented initializer init(nibName:bundle:)

What Causes the Crash and Why You Can't Solve It

The crash is caused by a bug in Cocoa. Unknown to you, init(style:) itself calls init(nibName:bundle:). And it calls it on self. That's you - MyTableViewController. But MyTableViewController has no implementation of init(nibName:bundle:). And does not inherit init(nibName:bundle:), either, because you already provided a designated initializer, thus cutting off inheritance.

Your only solution would be to implement init(nibName:bundle:). But you can't, because that implementation would require you to set the instance property greeting - and you don't know what to set it to.

The Simple Solution

The simple solution - almost too simple, which is why it is so difficult to think of - is: don't subclass UITableViewController. Why is this a reasonable solution? Because you never actually needed to subclass it in the first place. UITableViewController is a largely pointless class; it doesn't do anything for you that you can't do for yourself.

So, now we're going to rewrite our class as a UIViewController subclass instead. We still need a table view as our view, so we'll create it in loadView, and we'll hook it up there as well. Changes are marked as starred comments:

class MyViewController: UIViewController, UITableViewDelegate, UITableViewDataSource { // *
    let greeting : String
    weak var tableView : UITableView! // *
    init(greeting:String) {
        self.greeting = greeting
        super.init(nibName:nil, bundle:nil) // *
    }
    required init(coder aDecoder: NSCoder) {
        fatalError("init(coder:) has not been implemented")
    }
    override func loadView() { // *
        self.view = UITableView(frame: CGRectZero, style: .Plain)
        self.tableView = self.view as! UITableView
        self.tableView.delegate = self
        self.tableView.dataSource = self
    }
}

Also you'll want, of course, to add the minimal required data source methods. We now instantiate our class like this:

let tvc = MyViewController(greeting:"Hello there")

Our project compiles and runs without a hitch. Problem solved!

An Objection - Not

You might object that by not using UITableViewController we have lost the ability to get a prototype cell from the storyboard. But that is no objection, because we never had that ability in the first place. Remember, our hypothesis is that we are subclassing and calling our own subclass's initializer. If we were getting the prototype cell from the storyboard, the storyboard would be instantiating us by calling init(coder:) and the problem would never have arisen in the first place.

like image 138
matt Avatar answered Oct 17 '22 00:10

matt


Xcode 6 Beta 5

It appears that you can no longer declare a no-argument convenience initializer for a UITableViewController subclass. Instead, you need to override the default initializer.

class TestViewController: UITableViewController {
    override init() {
        // Overriding this method prevents other initializers from being inherited.
        // The super implementation calls init:nibName:bundle:
        // so we need to redeclare that initializer to prevent a runtime crash.
        super.init(style: UITableViewStyle.Plain)
    }

    // This needs to be implemented (enforced by compiler).
    required init(coder aDecoder: NSCoder!) {
        // Or call super implementation
        fatalError("NSCoding not supported")
    }

    // Need this to prevent runtime error:
    // fatal error: use of unimplemented initializer 'init(nibName:bundle:)'
    // for class 'TestViewController'
    // I made this private since users should use the no-argument constructor.
    private override init(nibName nibNameOrNil: String!, bundle nibBundleOrNil: NSBundle!) {
        super.init(nibName: nibNameOrNil, bundle: nibBundleOrNil)
    }
}
like image 20
Nick Snyder Avatar answered Oct 17 '22 01:10

Nick Snyder


Props to matt for a great explanation. I've made use of both matt's and @Nick Snyder's solutions, however I ran into a case in which neither would quite work, because I needed to (1) initialize let fields, (2) use init(style: .Grouped) (without getting a runtime error), and (3) use the built-in refreshControl (from UITableViewController). My workaround was to introduce an intermediate class MyTableViewController in ObjC, then use that class as the base of my table view controllers.

MyTableViewController.h

#import <UIKit/UIKit.h>
// extend but only override 1 designated initializer
@interface MyTableViewController : UITableViewController
- (instancetype)initWithStyle:(UITableViewStyle)style NS_DESIGNATED_INITIALIZER;
@end

MyTableViewController.m:

#import "MyTableViewController.h"
// clang will warn about missing designated initializers from
// UITableViewController without the next line.  In this case
// we are intentionally doing this so we disregard the warning.
#pragma clang diagnostic ignored "-Wobjc-designated-initializers"
@implementation MyTableViewController
- (instancetype)initWithStyle:(UITableViewStyle)style {
    return [super initWithStyle:style];
}
@end

Add the following to Project's Bridging-Header.h

#import "MyTableViewController.h"

Then use in swift. Example: "PuppyViewController.swift":

class PuppyViewController : MyTableViewController {
    let _puppyTypes : [String]
    init(puppyTypes : [String]) {
        _puppyTypes = puppyTypes // (1) init let field (once!)
        super.init(style: .Grouped) // (2) call super with style and w/o error
        self.refreshControl = MyRefreshControl() // (3) setup refresh control
    }
    // ... rest of implementation ...
}
like image 3
ɲeuroburɳ Avatar answered Oct 17 '22 00:10

ɲeuroburɳ