如果你想在纯swift中做到这一点，最好的方法是提供一个默认实现，特别是如果你返回一个swift类型，例如struct与swift类型

例子:

struct magicDatas {
    var damagePoints : Int?
    var manaPoints : Int?
}

protocol magicCastDelegate {
    func castFire() -> magicDatas
    func castIce() -> magicDatas
}

extension magicCastDelegate {
    func castFire() -> magicDatas {
        return magicDatas()
    }

    func castIce() -> magicDatas {
        return magicDatas()
    }
}

然后你可以实现协议而不定义每个func

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

您需要在每个方法之前添加可选关键字。 但是，请注意，要使其工作，您的协议必须用@objc属性标记。 这进一步表明该协议将适用于类，但不适用于结构。

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

为了说明安托万回答的机制:

protocol SomeProtocol {
    func aMethod()
}

extension SomeProtocol {
    func aMethod() {
        print("extensionImplementation")
    }
}

class protocolImplementingObject: SomeProtocol {

}

class protocolImplementingMethodOverridingObject: SomeProtocol {
    func aMethod() {
        print("classImplementation")
    }
}

let noOverride = protocolImplementingObject()
let override = protocolImplementingMethodOverridingObject()

noOverride.aMethod() //prints "extensionImplementation"
override.aMethod() //prints "classImplementation"

如何创建可选和必需的委托方法。

@objc protocol InterViewDelegate:class {

    @objc optional func optfunc()  //    This is optional
    func requiredfunc()//     This is required 

}