在Swift 3.0中

@objc protocol CounterDataSource {
    @objc optional func increment(forCount count: Int) -> Int
    @objc optional var fixedIncrement: Int { get }
}

这会节省你的时间。

1. 使用默认实现(首选)。

protocol MyProtocol {
    func doSomething()
}

extension MyProtocol {
    func doSomething() {
        /* return a default value or just leave empty */
    }
}

struct MyStruct: MyProtocol {
    /* no compile error */
}

优势

不涉及Objective-C运行时(好吧，至少没有显式地)。这意味着你可以使结构，枚举和非nsobject类符合它。此外，这意味着您可以利用强大的泛型系统。 当遇到符合这种协议的类型时，您总是可以确保满足所有需求。它总是具体实现或默认实现。这就是“接口”或“契约”在其他语言中的表现。

缺点

For non-Void requirements, you need to have a reasonable default value, which is not always possible. However, when you encounter this problem, it means that either such requirement should really have no default implementation, or that your you made a mistake during API design. You can't distinguish between a default implementation and no implementation at all, at least without addressing that problem with special return values. Consider the following example: protocol SomeParserDelegate { func validate(value: Any) -> Bool } If you provide a default implementation which just returns true — it's fine at the first glance. Now, consider the following pseudo code: final class SomeParser { func parse(data: Data) -> [Any] { if /* delegate.validate(value:) is not implemented */ { /* parse very fast without validating */ } else { /* parse and validate every value */ } } } There's no way to implement such an optimization — you can't know if your delegate implements a method or not. Although there's a number of different ways to overcome this problem (using optional closures, different delegate objects for different operations to name a few), that example presents the problem clearly.

2. 使用@objc可选。

@objc protocol MyProtocol {
    @objc optional func doSomething()
}

class MyClass: NSObject, MyProtocol {
    /* no compile error */
}

优势

不需要缺省实现。你只需要声明一个可选方法或变量就可以了。

缺点

它要求所有符合要求的类型都与Objective-C兼容，从而严重限制了协议的功能。这意味着，只有继承自NSObject的类才能符合这样的协议。没有结构，没有枚举，没有关联类型。 必须始终通过可选调用或检查符合类型是否实现了可选方法来检查是否实现了可选方法。如果经常调用可选方法，这可能会引入大量样板文件。

在协议中定义函数并为该协议创建扩展，然后为您想要作为可选使用的函数创建空实现。

下面是一个使用委托模式的具体示例。

设置您的协议:

@objc protocol MyProtocol:class
{
    func requiredMethod()
    optional func optionalMethod()
}

class MyClass: NSObject
{
    weak var delegate:MyProtocol?

    func callDelegate()
    {
        delegate?.requiredMethod()
        delegate?.optionalMethod?()
    }
}

将委托设置为类并实现协议。请注意，不需要实现可选方法。

class AnotherClass: NSObject, MyProtocol
{
    init()
    {
        super.init()

        let myInstance = MyClass()
        myInstance.delegate = self
    }

    func requiredMethod()
    {
    }
}

重要的一点是，可选方法是可选的，在调用时需要“?”。提到第二个问号。

delegate?.optionalMethod?()

由于有一些关于如何使用可选修饰符和@objc属性来定义可选需求协议的答案，我将给出一个如何使用协议扩展定义可选协议的示例。

下面的代码是Swift 3.*。

/// Protocol has empty default implementation of the following methods making them optional to implement:
/// `cancel()`
protocol Cancelable {

    /// default implementation is empty.
    func cancel()
}

extension Cancelable {

    func cancel() {}
}

class Plane: Cancelable {
  //Since cancel() have default implementation, that is optional to class Plane
}

let plane = Plane()
plane.cancel()
// Print out *United Airlines can't cancelable*

请注意，Objective-C代码不能调用协议扩展方法，更糟糕的是Swift团队不会修复它。https://bugs.swift.org/browse/SR-492

这里的其他答案涉及将协议标记为“@objc”，在使用swift特定类型时不起作用。

struct Info {
    var height: Int
    var weight: Int
} 

@objc protocol Health {
    func isInfoHealthy(info: Info) -> Bool
} 
//Error "Method cannot be marked @objc because the type of the parameter cannot be represented in Objective-C"

为了声明能在swift中很好地工作的可选协议，将函数声明为变量而不是func。

protocol Health {
    var isInfoHealthy: (Info) -> (Bool)? { get set }
}

然后实现如下协议

class Human: Health {
    var isInfoHealthy: (Info) -> (Bool)? = { info in
        if info.weight < 200 && info.height > 72 {
            return true
        }
        return false
    }
    //Or leave out the implementation and declare it as:  
    //var isInfoHealthy: (Info) -> (Bool)?
}

然后可以使用“?”来检查函数是否已经实现

func returnEntity() -> Health {
    return Human()
}

var anEntity: Health = returnEntity()

var isHealthy = anEntity.isInfoHealthy(Info(height: 75, weight: 150))? 
//"isHealthy" is true