Swift仿制药推迟了问题

Question

我正在努力做到这一点,但我遇到了一些麻烦

这是CustomProtocol

protocol CustomProtocol {

}

SubCustomProtocol

protocol SubCustomProtocol: CustomProtocol {

}

SubCustomProtocolImplementation

class SubCustomProtocolImplementation: SubCustomProtocol {

}

这是CustomClass

class CustomClass<P: CustomProtocol> {

    var customProtocol: P?

    func doSomething() {

    } 

}

SubCustomClass

class SubCustomClass<P: SubCustomProtocol>: CustomSubClass {

}

和我的BaseViewController

class BaseViewController<P: CustomProtocol, T: CustomClass<P>>: UIViewController {

    var foo: T!

    override func viewDidLoad() {
        super.viewDidLoad()
        foo?.doSomething()
    }
}

我的ViewController

class ViewController<P: SubCustomProtocolImplementation, T: SubCustomClass<P>>: BaseViewController<P,T> {

    override func viewDidLoad() {
        super.viewDidLoad()
    }
}

在我称之为foo？.doSomething()的行中,它说'T'不是'CustomClass <'P'>'的子类型,我不知道我做错了什么

在ViewController声明中,它说"BaseViewController要求T继承自CustomClass <'P'>"

希望你能帮我!

Answer 1

如果您想指定您的 foo var 类型，CustomClass<P>您应该按照以下方式执行。

class ViewController<P: CustomProtocol>: UIViewController {

    var foo: CustomClass<P>?

    override func viewDidLoad() {
        super.viewDidLoad()
        foo?.doSomething()
    }
}