Swift 2.3到Swift 3.0对成员“加入”获取歧义引用错误

Question

我正在研究将Swift 2.3转换为Swift 3.0;

Swift 2.3代码：

extension ContextDidSaveNotification: CustomDebugStringConvertible {
    public var debugDescription: String {
        var components = [notification.name]
        components.append(managedObjectContext.description)
        for (name, set) in [("inserted", insertedObjects), ("updated", updatedObjects), ("deleted", deletedObjects)] {
            let all = set.map { $0.objectID.description }.joinWithSeparator(", ")
            components.append("\(name): {\(all)}")
        }
        return components.joinWithSeparator(" ")
    }
}

Swift 3.0代码：

extension ContextDidSaveNotification: CustomDebugStringConvertible {
    public var debugDescription: String {
        var components = [notification.name]
        components.append(Notification.Name(rawValue: managedObjectContext.description))
        for (name, set) in [("inserted", insertedObjects), ("updated", updatedObjects), ("deleted", deletedObjects)] {
            let all = set.map { $0.objectID.description }.joined(separator: ", ")
            components.append(Notification.Name(rawValue: "\(name): {\(all)}"))
        }
        return components.joined(separator: " ")
    }
}

但是我得到了一个错误：在Swift 3.0代码中，有误地对成员'joined（）'进行了归还。

如何解决这个问题呢？我做了很多研究，但找不到可行的解决方案。

谢谢

Answer 1

joined(separator:)声明为Array<String>，不是Array<Notification.Name>：

// somewhere in standard library
extension Array where Element == String {

    public func joined(separator: String = default) -> String
}

正如@vadian指出的，Notification.Name它不等于String，因此您需要先转换数组。这应该工作：

components
  .map({ $0.rawValue })
  .joined(separator: " ")