我仍然无法理解 
 Swift中泛型的一些微妙之处.我定义了以下类型: 
  
  
 
protocol SomeProtocol {
    func setValue(value: Int)
}
class ProtocolLabel : UILabel,SomeProtocol {
    func setValue(value: Int) {
    }
}
class ProtocolImageView : UIImageView,SomeProtocol {
    func setValue(value: Int) {
    }
} 
 viewForValue(2)
现在我定义了以下功能.我希望T是一个符合协议SomeProtocol的UIView.
func viewForValue<T where T: SomeProtocol,T: UIView>(param: Int) -> UIView {
    var someView: T
    if param > 0 {
        someView = ProtocolLabel() as T
    } else {
        someView = ProtocolImageView() as T
    }
    someView.setValue(2)
    someView.frame = CGRectZero
    return someView
} 
 但是,当我执行代码时,我收到以下编译错误:
viewForValue(2) // <-- Type 'UIView' does not conform to protocol 'SomeProtocol'
似乎在where子句中我不能指定不实现协议的类.这是为什么?
提前致谢.
解决方法
 viewForValue应该返回一个继承自UIView并实现SomeProtocol的类. 
 
您已经定义了两个没有直接关系的类 – 它们只是从UIView继承并实现SomeProtocol.
        您已经定义了两个没有直接关系的类 – 它们只是从UIView继承并实现SomeProtocol.
当函数必须确定返回类型时,两个类继承的直接具体类型是UIView,这就是viewForValue返回的内容.
为了解决这个问题,你必须通过创建一个继承自UIView并实现SomeProtocol的第三类来创建两个类之间的直接和具体的关系:
protocol SomeProtocol {
    func setValue(value: Int)
}
class SomeClass: UIView,SomeProtocol {
    func setValue(value: Int) {
    }
}
class SomeSubclass : SomeClass {
}
class SomeOtherSubclass : SomeClass {
}
func viewForValue<T where T: SomeProtocol,T: SomeClass>(param: Int) -> T {
    var someView: T
    if param > 0 {
        someView = SomeSubclass() as T
    } else {
        someView = SomeOtherSubclass() as T
    }
    someView.setValue(2)
    someView.frame = CGRectZero
    return someView
}
viewForValue(2) 
 附录:阅读下面的OP评论,目的是动态实例化从UIView继承的现有UIKit类.因此,建议的解决方案不适用.
我认为通过实现SomeProtocol来扩展UIView应该可行:
protocol SomeProtocol {
    func setValue(value: Int)
}
extension UIView : SomeProtocol {
    func setValue(value: Int) {
    }
}
func viewForValue<T where T: SomeProtocol,T: UIView>(param: Int) -> UIView {
    var someView: T
    if param > 0 {
        someView = UILabel() as T
    } else {
        someView = UIImageView() as T
    }
    someView.setValue(2)
    someView.frame = CGRectZero
    return someView
} 
 但看起来有一个编译器错误.操场上的此代码显示一条消息,指出:
Communication with the playground service was interrupted unexpectedly.
The playground service “com.apple.dt.Xcode.Playground” may have generated a crash log.
而在iOS应用程序中,由于分段错误,编译失败11.