I just joined a project that has a lot of existing code. The previous programmer was perhaps unfamiliar with Swift or began development in the early stages of the Swift language. They seemed to be using the if let statement in an odd way. They seemed to want to use the statement as a if is let. Before I edit the code I would like to know if there is any valid use for this:
// In JSON parser
if value is String, let string = value as? String {
document.createdBy = string
}
First checking if value is of type String seems redundant to me. Doesn't Swift check for this in the let string = value as? String portion of the statement?
QUESTION
Why would this need to be checked twice? Or would there be a reason for this?
You're correct, this is redundant. If value is not a string, then value as? String would return nil, and the conditional binding would fail.
To check the type, and not use the casted result:
if value is String {
// Do something that doesn't require `value` as a string
}
To check the type and use the result:
if let value = value as? String { // The new name can shadow the old name
document.createdBy = value
}
Doing both makes no sense.
Related
Edit
let disucssionMessageTimestampKey = DiscussionMessage.CodingKeys.messageTimestamp.stringValue gives an error:
'CodingKeys' is inaccessible due to 'private' protection level
I have a message structure defined like this:
struct DiscussionMessage: Codable {
let message, userCountryCode, userCountryEmoji, userName, userEmailAddress: String
let messageTimestamp: Double
let fcmToken, question, recordingUrl, profilePictureUrl: String?
}
I want to define a variable disucssionMessageTimestampKey whose value will be messageTimestamp. I want to use disucssionMessageTimestampKey variable in the following query:
messagesReference.queryOrdered(byChild: "messageTimestamp").queryStarting(atValue: NSDate().timeIntervalSince1970).observe(.childAdded)
So that I don't have to hardcode the string value ("messageTimestamp") of the variable name.
Now I know I could just do let disucssionMessageTimestampKey: String = "messageTimestamp". But this is again prone to errors. So I was wondering if there was a way that I could get the string value messageTimestamp without having to define it anywhere.
By something like this (I know this won't work but just to give an idea of what I am looking for)
let disucssionMessageTimestampKey: String = String(describing: DiscussionMessage.messageTimestamp) // Will store disucssionMessageTimestampKey = "messageTimestamp"
Also, would it be possible to completely define the key values first as strings and then use those as variable names in the actual codable object? I.e. first define let disucssionMessageTimestampKey: String = "messageTimestamp", and then use the variable disucssionMessageTimestampKey to define what the property (messageTimestamp) of the codable object should be called. (This is low priority but curious and seems related to the question at hand)
I have tried
self.adc_role_id = String(res["adc_role_id"])
self.adc_role_id = "\(res["adc_role_id']"
self.adc_role_id = (\(res["adc_role_id"] as? String)!
but still get
Could not cast value of type '__NSCFNumber' to 'NSString'
I added the dump of res[4] below
As new as I am to Swift, I don't know anything else to try
In Swift 4, the String initializer requires the describing: argument label.
I don't know if this will solve your problem, but your first line of code should be written:
self.adc_role_id = String(describing: res["adc_role_id"])
In your screenshot we can see that res["adc_role_id"] is an NSNumber.
To transform an NSNumber to a String you should use its stringValue property.
And since a dictionary gives an Optional, you should use optional binding to safely unwrap it.
Example:
if let val = res["adc_role_id"] {
self.adc_role_id = val.stringValue
}
You could also, if you want, use string interpolation instead of the property:
if let val = res["adc_role_id"] {
self.adc_role_id = "\(val)"
}
but I think using the property is more relevant.
If for some reason the compiler complains about the type of the content, cast it:
if let val = res["adc_role_id"] as? NSNumber {
self.adc_role_id = val.stringValue
}
Note that you should not use String(describing:) because this initializer will try to represent the string in many ways, and some of them will give inaccurate and unexpected results (for example, if String(describing:) resolves to use the debugDescription property, as explained in the documentation, you may get a totally different string than the one you want).
It's also worth noting that using String(describing:) with an optional value such as your dictionary will resolve to a wrong string: String(describing: res["adc_role_id"]) will give Optional(yourNumber)! This is why Mike's answer is wrong. Be careful about this. My advice is to avoid using String(describing:) altogether unless for debugging purposes.
The error message is clear and the dump is clear, too.
The value is not a String, it's an Int(64) wrapped in NSNumber
Optional bind the value directly to Int (NSNumber is implicit bridged to Int) and use the String initializer.
if let roleID = res["adc_role_id"] as? Int {
self.adc_role_id = String(roleID)
}
Please conform to the naming convention that variable names are camelCased rather than snake_cased
I have a bit of code to get a string out of userDefaults:
if let userString = (userDefaults.objectForKey("user")) {
userTextField.stringValue = userString as! String
}
First, I have to see if the optional is not nil. Then I have to cast it as a string from AnyObject.
Is there a better way of doing this? maybe a one liner?
Note that your forced cast as! String will crash if a default value for the key "user" exists, but
is not a string. Generally, you can combine optional binding (if let) with an optional cast (as?):
if let userString = userDefaults.objectForKey("user") as? String {
// ... default value for key exists and is a string ...
userTextField.stringValue = userString
}
But actually NSUserDefaults has a dedicated method for that purpose:
if let userString = userDefaults.stringForKey("user") {
// ... default value for key exists and is a string ...
userTextField.stringValue = userString
}
If you want to assign a default string in the case that
the default does not exist, or is not a string, then use the
nil-coalescing operator ??, as demonstrated in
Swift issue with nil found while unwrapping an Optional value NSDefautlts, e.g.:
userTextField.stringValue = userDefaults.stringForKey("user") ?? "(Unknown)"
For the special case NSUserDefaults the best – and recommended – way is to use always non-optional values.
First register the key / value pair in AppDelegate as soon as possible but at least before using it.
let defaults = NSUserDefaults.standardUserDefaults()
let defaultValues = ["user" : ""]
defaults.registerDefaults(defaultValues)
The benefit is you have a reliable default value of an empty string until a new value is saved the first time. In most String cases an empty string can be treated as no value and can be easily checked with the .isEmpty property
Now write just
userTextField.stringValue = userDefaults.stringForKey("user")!
Without arbitrary manipulation of the defaults property list file the value is guaranteed to be never nil and can be safely unwrapped, and when using stringForKey there is no need for type casting.
Another way that i like much to clean this up is to do each of your checks
first, and exit if any aren’t met. This allows easy understanding of what
conditions will make this function exit.
Swift has a very interesting guard statements which can also be used to avoid force unwrap crashes like :
guard let userString = userDefaults.objectForKey("user") as? String else {
// userString var will accessible outside the guard scope
return
}
userTextField.stringValue = userString
Using guards you are checking for bad cases early, making your
function more readable and easier to maintain. If the condition is not
met, guard‘s else statement is run, which breaks out of the function.
If the condition passes, the optional variable here is automatically
unwrapped for you within the scope that the guard statement was
called.
how can i convert any object type to a string?
let single_result = results[i]
var result = ""
result = single_result.valueForKey("Level")
now i get the error: could not assign a value of type any object to a value of type string.
and if i cast it:
result = single_result.valueForKey("Level") as! String
i get the error:
Could not cast value of type '__NSCFNumber' (0x103215cf0) to 'NSString' (0x1036a68e0).
How can i solve this issue?
You can't cast any random value to a string. A force cast (as!) will fail if the object can't be cast to a string.
If you know it will always contain an NSNumber then you need to add code that converts the NSNumber to a string. This code should work:
if let result_number = single_result.valueForKey("Level") as? NSNumber
{
let result_string = "\(result_number)"
}
If the object returned for the "Level" key can be different object types then you'll need to write more flexible code to deal with those other possible types.
Swift arrays and dictionaries are normally typed, which makes this kind of thing cleaner.
I'd say that #AirSpeedVelocity's answer (European or African?) is the best. Use the built-in toString function. It sounds like it works on ANY Swift type.
EDIT:
In Swift 3, the answer appears to have changed. Now, you want to use the String initializer
init(describing:)
Or, to use the code from the question:
result = single_result.valueForKey("Level")
let resultString = String(describing: result)
Note that usually you don't want valueForKey. That is a KVO method that will only work on NSObjects. Assuming single_result is a Dictionary, you probably want this syntax instead:
result = single_result["Level"]
This is the documentation for the String initializer provided here.
let s = String(describing: <AnyObject>)
Nothing else is needed. This works for a diverse range of objects.
The toString function accepts any type and will always produce a string.
If it’s a Swift type that implements the Printable protocol, or has overridden NSObject’s description property, you’ll get whatever the .description property returns. In the case of NSNumber, you’ll get a string representation of the number.
If it hasn’t, you’ll get a fairly unhelpful string of the class name plus the memory address. But most standard classes, including NSNumber, will produce something sensible.
import Foundation
class X: NSObject {
override var description: String {
return "Blah"
}
}
let x: AnyObject = X()
toString(x) // return "Blah"
"\(x)" // does the same thing but IMO is less clear
struct S: Printable {
var description: String {
return "asdf"
}
}
// doesn't matter if it's an Any or AnyObject
let s: Any = S()
toString(s) // reuturns "asdf"
let n = NSNumber(double: 123.45)
toString(n) // returns "123.45"
n.stringValue // also works, but is specific to NSNumber
(p.s. always use toString rather than testing for Printable. For one thing, String doesn’t conform to Printable...)
toString() doesn't seem to exist in Swift 3 anymore.
Looks like there's a failable initializer that will return the passed in value's description.
init?(_ description: String)
Docs here https://developer.apple.com/reference/swift/string/1540435-init
I have an "if let" statement that is being executed, despite the "let" part being nil.
if let leftInc : Double? = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!]! {
println(leftInc)
let valueString : String = formatter.stringFromNumber(NSNumber(double: leftInc!))!
self.leftIncisorTextField?.text = valueString
self.btnLeftIncisor.associatedLabel?.text = valueString
}
// self.analysis.inputs is a Dictionary<String, Double?>
The inputs dictionary holds information entered by the user - either a number, or nil if they haven't entered anything in the matching field yet.
Under the previous version of Swift, the code was written as this:
if let leftInc : Double? = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!]?? {
and worked correctly.
I saw a similar question here, but in that instance the problem seemed to be the result of using Any?, which is not the case here.
Swift 2.2
In your if let you define another optional, that's why nil is a legitimate case. if let is intended mainly to extract (maybe) non optional value from an optional.
You might try:
if let leftInc : Double = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!].flatMap ({$0}) {
// leftInc is not an optional in this scope
...
}
Anyway I'd consider to not do it as a one liner but take advantage of guard case. Just in order to enhance readability. And avoid bang operator (!).
The if-let is for unwrapping optionals. You are allowing nil values by setting the type to an optional Double.
The if statement should be:
if let leftInc = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!] as? Double{
...
}
This will attempt to get an object out of inputs, if that fails it returns nil and skips it. If it does return something it will attempt to convert it to a Double. If that fails it skips the if statement as well.
if inputs is a dictionary like [Something:Double] then you don't need the last as? Double as indexing the dictionary will return a Double?
I recommend reading the swift book on optional chaining.
You could break it down further -
if let optionalDouble = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!], leftInc = optionalDouble {
....
}
as your dictionary has optional values - this way of writing it might make it clearer what's going on
if let k = dict["someKey"]{}, dict["someKey"] will be an object of type Any
this can bypass a nill
So do a typecast to get it correct like if let k = dict["someKey"] as! String {}