創建擴展導致「錯誤類型」是無法轉換爲「NSError」

Question

我一直在玩NSDecimalNumber，當我創建了一個擴展，突然間，我收到此錯誤：

'ErrorType' is not convertible to 'NSError'; did you mean to use 'as!' to force downcast?

但是，當我使用as!的建議，我得到了警告：

Forced cast from 'NSMutableDictionary!' to '[NSObject : AnyObject]' always succeeds; did you mean to use 'as'?

最終轉變爲as?是給我：

Conditional cast from 'ErrorType' to 'NSError' always succeeds

我創建的擴展名是：

extension String { 

    var decimalValue: NSDecimalNumber? { 
     let number = NSDecimalNumber(string: self) 

     return number == NSDecimalNumber.notANumber() ? nil : number 
    } 
} 

導致該錯誤代碼是：

lazy var persistentStoreCoordinator: NSPersistentStoreCoordinator = { 
    let coordinator = NSPersistentStoreCoordinator(managedObjectModel: self.managedObjectModel) 
    let url = self.applicationDocumentsDirectory.URLByAppendingPathComponent(self.storeFile) 
    var failureReason = "There was an error creating or loading the application's saved data." 

    do { 
     try coordinator.addPersistentStoreWithType(NSSQLiteStoreType, configuration: nil, URL: url, options: nil) 
    } catch { 
     var dict = [String: AnyObject]() 
     dict[NSLocalizedDescriptionKey] = "Failed to initialize the application's saved data" 
     dict[NSLocalizedFailureReasonErrorKey] = failureReason 
---> dict[NSUnderlyingErrorKey] = error as NSError <--- 

     let error = NSError(domain: "YOUR_ERROR_DOMAIN", code: 9999, userInfo: dict) 

     NSLog("Unresolved error \(error), \(error.userInfo)") 
     abort() 
    } 

    return coordinator 
}() 

當我改擴建這個錯誤將會消失：

var decimalValue: NSDecimalNumber? { 
    return NSDecimalNumber(string: self) 
} 

如何我應該做這個擴展沒有得到這個惱人的警告/錯誤？

Answer 1

我不知道爲什麼它只是因爲您將擴展名添加到String而引起警告，但我們可以照顧警告！

由於this SO post answer suggests，我們可以這樣做：

lazy var persistentStoreCoordinator: NSPersistentStoreCoordinator = { 
    .... 
    do { 
     try coordinator.addPersistentStoreWithType(NSSQLiteStoreType, configuration: nil, URL: url, options: nil) 
    } catch let error as NSError { 
     .... 
     dict[NSUnderlyingErrorKey] = error 
     let wrappedError = NSError(domain: "YOUR_ERROR_DOMAIN", code: 9999, userInfo: dict) 
     .... 
     NSLog("Unresolved error \(wrappedError), \(wrappedError.userInfo)") 
     abort() 
    } catch { 
     // dummy 
    } 

    return coordinator 
}() 

關鍵因素是catch錯誤就像catch let error as NSError

Answer 2

下面是答案其實該消息是很清楚的。 ErrorType是一個空協議，它已被NSError採用。因此，沒有必要「保護」或「如果讓」再次。你可以簡單地說

let underlyingError = error as NSError 

然後警告就會消失。