When a subclass overrides a method, how can we ensure at compile time that the superclass's method implementation is called?

Question

The classic example is:

- (void)viewDidLoad {
    [super viewDidLoad]; // Subclasses sometimes forget this line

    // Subclass\'s implementation goes here
         


        

Answer 1

If we're talking about custom classes, you can add the following to your superclass's method declaration:

__attribute__((objc_requires_super));

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And if you want to ensure that all of your UIViewController subclasses call a method like [super viewDidLoad];, you could subclass UIViewController something like this:

@interface BaseViewController : UIViewController

- (void)viewDidLoad __attribute__((objc_requires_super));

// per Scott's excellent comment:
- (void)viewWillAppear:(BOOL)animated NS_REQUIRES_SUPER;

@end

---

@implementation BaseViewController

- (void)viewDidLoad {
    [super viewDidLoad];
}

- (void)viewWillAppear:(BOOL)animated {
    [super viewWillAppear:animated];
}

@end

And then just subclass BaseViewController throughout your project, rather than subclassing UIViewController.

Any subclass of BaseViewController which implements viewDidLoad and does not call [super viewDidLoad]; (which in turn calls UIViewController's viewDidLoad) will throw a warning.

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EDIT: I've edited the answer to include an example of NS_REQUIRES_SUPER, per Scott's excellent comment. The two examples (viewDidLoad and viewWillAppear:) are functionally equivalent. Though I imagine NS_REQUIRES_SUPER probably will autocomplete for you. I'll likely begin using this macro myself in the future.