@protocol I <NSObject>
-(void) f;
-(void) g;
@end
@interface C : NSObject <I>
{
id<I> i;
}
-(void) toA;
-(void) toB;
@end
I am using this code, Here I am using a protocol I, I want to know what is the meaning of ( id< I > i ) and working of this.
Declaration:
MyClass<MyProtocol> * myVar;
means that class of myVar is one of MyClass descendants and additionaly implements methods of MyProtocol protocol. Read more about protocols here.
And in your code
id<I> i;
means that "i" is of id type - i.e. it can be object of any class (read about id here), but is also conforms to protocol "I".
In you sample class of your "i" must implement
-(void) f;
-(void) g;
methods.
And you can use expressions like:
[i f];
[i g];
It means that the iVar i of any type (thus id), conforms to protocol I.
You declared i as an Objective-C object of type id that conforms to your protocol I.
Read more about it here: http://unixjunkie.blogspot.com/2008/03/id-vs-nsobject-vs-id.html
i is an object. It could be any class, as long as that class conforms to the protocol I. That means that class has instance methods f and g that take no arguments and return nothing. That class also must formally adopt protocol I. Class C actually adopts protocol I. Inside a method of class C you could write:
i = self;
Though that might not usually be very useful. More likely some other class adopts protocol I too, or at least you would assign a different instance of C to i, like
i = [[C alloc] init];
id refers to a variable that adheres to the protocol I. In this case, you would then need i to be be a variable whose class implements protocol I.
You probably want to read up on the delegates here: How to use custom delegates in Objective-C
The id i is your delegate property. When you use it. You can call one of the delegate methods you described in your protocol, aka method f and g. In the class that comforms to the protocol you given shape to these methods.
