ios - 如何返回基于类类型的对象？

Question

这是我出于测试目的需要的:

class AssemblerMock: Assemblerable {
    func resolve<Service>(_ serviceType: Service.Type) -> Service? {
        return Service.init() //doesnt work, need to return non nil value here.
    }
}

Answer 1

它有一些变通方法:您需要创建一个协议(protocol)，我们称它为 Initable:

protocol Initable {
    init()
}

然后，您的 resolve-Template-Method 应该要求 Service 为 Initable:

func resolve<Service>(_ serviceType: Service.Type) -> Service where Service:Initable {
    return Service.init()
}

在使用它之前，您还需要为您可能想要解析的所有类型创建一个扩展:

extension Int : Initable {
    // No implementation possible/needed, because `init` already exits in struct Int
} 

然后调用它:

let am = AssemblerMock()
let i = am.resolve(Int.self)
print (i)   // Prints "0" because this is the default Integer value

备注:我把返回类型设为返回Service而不是Service?，不过这里无所谓。如果要支持可失败初始化器(init?)，则需要修改返回类型以及Initable 协议(protocol):

protocol Initable {
    init?()
}

extension Int : Initable {} 

class FooFailing : Initable {
    required init?() {
        return nil
    }
}

class AssemblerMock {
    func resolve<Service>(_ serviceType: Service.Type) -> Service? where Service:Initable {
        return Service.init()
    }
}

let am = AssemblerMock()
let i = am.resolve(Int.self)
print (i)   // Optional(0)
let foo = am.resolve(FooFailing.self)
print (foo) // nil