在 Objective-C 中将类的实例转换为 @protocol
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Cast an instance of a class to a @protocol in Objective-C
提问by Ford
I have an object (a UIViewController) which may or may not conform to a protocol I've defined.
我有一个对象（一个 UIViewController），它可能符合也可能不符合我定义的协议。
I know I can determine if the object conforms to the protocol, then safely call the method:
我知道我可以确定对象是否符合协议，然后安全地调用该方法：
if([self.myViewController conformsToProtocol:@protocol(MyProtocol)]) {
[self.myViewController protocolMethod]; // <-- warning here
}
However, XCode shows a warning:
但是，XCode 显示警告：
warning 'UIViewController' may not respond to '-protocolMethod'
What's the right way to prevent this warning? I can't seem to cast self.myViewControlleras a MyProtocolclass.
防止此警告的正确方法是什么？我似乎无法self.myViewController作为一个MyProtocol班级演员。
回答by Nick Forge
The correct way to do this is to do:
正确的方法是这样做：
if ([self.myViewController conformsToProtocol:@protocol(MyProtocol)])
{
UIViewController <MyProtocol> *vc = (UIViewController <MyProtocol> *) self.myViewController;
[vc protocolMethod];
}
The UIViewController <MyProtocol> *type-cast translates to "vc is a UIViewController object that conforms to MyProtocol", whereas using id <MyProtocol>translates to "vc is an object of an unknown class that conforms to MyProtocol".
该UIViewController <MyProtocol> *型铸造转换为“VC是一个UIViewController对象符合MyProtocol”，而使用id <MyProtocol>转换为“VC是一个未知类符合MyProtocol的目的”。
This way the compiler will give you proper type checking on vc- the compiler will only give you a warning if any method that's not declared on either UIViewControlleror <MyProtocol>is called. idshould only be used in the situation if you don't know the class/type of the object being cast.
这样，编译器将为您提供正确的类型检查vc- 编译器只会在任何未声明UIViewController或未<MyProtocol>调用的方法时向您发出警告。id如果您不知道正在转换的对象的类/类型，则应仅在这种情况下使用。
回答by Andy
You can cast it like this:
你可以这样投射：
if([self.myViewController conformsToProtocol:@protocol(MyProtocol)])
{
id<MyProtocol> p = (id<MyProtocol>)self.myViewController;
[p protocolMethod];
}
This threw me for a bit, too. In Objective-C, the protocol isn't the type itself, so you need to specify id(or some other type, such as NSObject) along with the protocol that you want.
这也让我有点受宠若惊。在 Objective-C 中，协议不是类型本身，因此您需要指定id（或某些其他类型，例如NSObject）以及您想要的协议。
