具有协议继承的objc不兼容指针类型

Question

I am running with Xcode version 5.1, iOS SDK version 7.1. Here are some sample declarations all in the same file:

@protocol A <NSObject>
@end

@protocol B <A>
@end

@interface SomeObject : NSObject <B>
@end

@interface SomeContainer : NSObject
- (id<A>)pop;
@end

Xcode generates a warning in the following code:

SomeContainer *container = [[SomeContainer alloc] init];
SomeObject *obj = [container pop]; // Warning: Initializing 'SomeObject *__strong' with and expression of incompatible type 'id<A>'

I know that typecasting will get rid of the warning, however, I don't understand why I need it. SomeObject must implement whatever is declared in protocol A. Is there something that I am missing? Any explanation would be greatly appreciated.

Answer 1

Probably this is because SomeObject is something more than just id<A> . It's like initialising an object of derived class with pointer to the base class. You can initialize an instance of id<A> with SomeObject but when you do it vice versa you're receiving a warning, which says :"Hey, buddy you've just assigned base to derived, ie something that is present in derived (SomeObject *) may be missed in base id<A> "

for example SomeObject may contain method foo which is not present in protocol A , but now after that assignment of yours if you send foo message to obj it may crash, because id<A> is not necessarily has this method. It can, if it's also an instance of SomeObject , but it can be anything, which NECESSARILY responds only to the methods declared in A .

I hope all this makes sense