Question

当我尝试调用存储的闭包时遇到以下错误。我尝试构建时遇到此错误：

＆＃39;（T） - ＆gt;虚空＆＃39;不可转换为＆＃39; @lvalue（T） - ＆gt;虚空＆＃39; （又名＆＃39; @lvalue（T） - ＆gt;（）＆＃39;）

public protocol DataSourceProtocol {
  associatedtype DataSourceItem

  func item(indexPath: IndexPath) -> DataSourceItem?
  func update<T>(sender : T)
}

public class AnyDataSourceSimple<T> : DataSourceProtocol {
  private var itemClosure : (IndexPath) -> T?
  private var updateClosure: (T) -> Void

  public init<I: DataSourceProtocol>(_ concrete: I) where I.DataSourceItem == T {
    self.itemClosure = concrete.item
    self.updateClosure = concrete.update
}

public func item(indexPath: IndexPath) -> T? {
    return itemClosure(indexPath)
}

public func update<T>(sender: T) {
    // '(T) -> Void' is not convertible to '@lvalue (T) -> Void' (aka '@lvalue (T) -> ()')
    updateClosure(sender)   
    print("update")
  }
}

这是否与协议中的通用函数定义有某种关系？

Answer 1

与评论中一样，通用函数的T与泛型类定义中的T是分开的。

要使它编译，你必须这样做，不确定这是否是你的意思

import Foundation

public protocol DataSourceProtocol {
    associatedtype DataSourceItem
    associatedtype UpdateSender

    func item(indexPath: IndexPath) -> DataSourceItem?
    func update(sender : UpdateSender)
}

public class AnyDataSourceSimple<T> : DataSourceProtocol {
    private var itemClosure : (IndexPath) -> T?
    private var updateClosure: (T) -> Void

    public init<I: DataSourceProtocol>(_ concrete: I) where I.DataSourceItem == T, I.UpdateSender == T {
        self.itemClosure = concrete.item
        self.updateClosure = concrete.update
    }

    public func item(indexPath: IndexPath) -> T? {
        return itemClosure(indexPath)
    }

    public func update(sender: T) {
        updateClosure(sender)   
        print("update")
    }
}

通用协议功能＆amp;快速升值

1 个答案: