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Home/ Questions/Q 7061869
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T04:31:34+00:00 2026-05-28T04:31:34+00:00

I want to define a protocol and create an easy, standard way to grab

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I want to define a protocol and create an easy, standard way to grab a ‘default’, shared implementation of said protocol – singleton style. Cocoa adhere’s to the following pattern:

[NSUserDefaults standardUserDefaults]
[NSNotificationCenter defaultCenter]

but in both cases, they have @interfaces at the bottom of the object hierarchy. I’m struggling with how to do this using @protocols. I can obviously create a class that has empty or simple method implementations – but in reality, what I want is a @protocol at the bottom of the hierarchy. I’ve tried something like:

@protocol ConfigurationManager <NSObject>

//...

@interface ConfigurationManagerFactory : NSObject

+ (id<ConfigurationManager>)sharedConfiguration;

@end

// ...

id<ConfigurationManger> config = [ConfigurationManagerFactory sharedConfiguration];
[config ...];

and it works – but I’m always having to explain how to use this and why I did it this way. Is there a way to conform to Cocoa’s syntax (calling convention) while still leveraging the value of @protocols?

As an aside, is there a reason why I wouldn’t want to use @protocols like this? The implementing @interface can still leverage categories and alternate implementations, etc – just like how instantiating an NSString usually leaves you with a class extending NSString.

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  1. Editorial Team
    Editorial Team
    2026-05-28T04:31:35+00:00Added an answer on May 28, 2026 at 4:31 am

    Here’s an idea: create your protocol and a class with the same name with a factory method that returns you the default implementation of the protocol:

    @protocol ConfigurationManager <NSObject> ...
    
    @interface ConfigurationManager : NSObject <ConfigurationManager> 
    +(ConfigurationManager *) defaultConfigurationManager;
    ...
    

    Other specialized implementations can then inherit from your base class.

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