Get a user-readable version of the class name in swift (in objc NSStringFromClass was fine)

Question

Is there an equivalent of NSStringFromClass in Swift that gives a user-readable version of the class name? I've tried to use it with a native Swift class I created, but as you can see, the result seems to be the compiler's internal representation of the class name:

println(NSStringFromClass(MyClass.self))

Result:

_TtC5test7MyClass

I've tried adding the @objc attribute to the class, and making it a subclass of NSObject, but it makes no difference. I've discovered that if I replace MyClass with an Objective-C class of the same name, and import this in the bridging header, it gives me 'MyClass', but this shouldn't be necessary.

Another option would be to make a protocol for this, which each class I want to use in this way must conform to:

protocol Nameable { class var className: String { get } }

However, it would be easier to have a built-in way to do this in the language.

NatashaTheRobot · Accepted Answer · 2016-01-15 00:00:55Z

64

You can now just do:

String(MyClass)

answered Jan 15, 2016 at 0:00

NatashaTheRobot

6,9294 gold badges34 silver badges27 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

fractious Over a year ago

This makes me unreasonably happy! Thanks Natasha robot.

shim Over a year ago

In Swift 3 this doesn't work anymore. The Swift 3 converter changed it to String(describing: MyClass) but that's not working either.

Apan Over a year ago

@shim have you tried String(describing: self) or just passing object let object = MyClass() String(describing: object)

ImpactZero · Accepted Answer · 2015-04-15 14:11:05Z

new format based on xcode 6 beta 5.

(project_name).(class_name)

func getName(classType:AnyClass) -> String { let classString = NSStringFromClass(classType.self) let range = classString.rangeOfString(".", options: NSStringCompareOptions.CaseInsensitiveSearch, range: Range<String.Index>(start:classString.startIndex, end: classString.endIndex), locale: nil) return classString.substringFromIndex(range!.endIndex) }

Latest 6.3 Xcode Swift 1.2

if you need an extension or you can put this on any common object:

public extension NSObject{ public class var nameOfClass: String{ return NSStringFromClass(self).componentsSeparatedByString(".").last! } public var nameOfClass: String{ return NSStringFromClass(self.dynamicType).componentsSeparatedByString(".").last! } }

this also works: NSStringFromClass(self).componentsSeparatedByString(".").last!
As @NatashaTheRobot states below. Just use String(MyClass) now.

Timonea · Accepted Answer · 2016-12-17 06:46:34Z

Swift 3

type(of: self) prints ModuleName.ClassName

String(describing: type(of: self)) prints ClassName

Community · Accepted Answer · 2020-06-20 09:12:55Z

At the moment, there's no reliable way to do this. See an Apple developer's comment on https://devforums.apple.com/thread/227425

Swift does not currently have much in the way of introspection.

There is some introspection machinery that is used for the playgrounds. I don't know if that is intended to be API.

Some Swift methods and variables can be examined using the Objective-C runtime's introspection. That's likely to be the best solution today.

Swift does have the notion of a metatype, somewhat analogous to the Class type in Objective C. You can find it using TypeName.self, e.g.:

class Foo { @required init() { } } var normalFoo : Foo = Foo() var fooType : Foo.Type = Foo.self; var fooFromMetatype : Foo = fooType();

Perhaps, by release time, metatypes will include more introspection abilities. I suggest filing a Radar feature request for this.

hnh · Accepted Answer · 2015-07-27 11:39:19Z

In Swift 2 beta 4 you can get to the information via the type object:

let className = "\(String.self)" // gives: "Swift.String"

or if you have an instance:

let s = "Hello World" let className = "\(s.dynamicType)" // gives: "Swift.String"

You get the Module.Class result, like:

Swift.String ABC.MyGenericClass<Swift.Int>

Funny enough the Type object returned does not seem to conform to the CustomStringConvertible protocol. Hence it has no 'description' property, though the "" pattern still does the right thing.

P.S.: Before b4 the same could be accomplished via reflect(obj.dynamicType).summary.

That's cool but until they have an explicit className() method somewhere in there I am a little wary. Summary might be the class name right now but there are no guarantees it'll stay that way.

Coach Roebuck · Accepted Answer · 2016-10-05 17:35:22Z

3

In Swift v3.0, this worked for me:

String.init(describing: self.self)

answered Oct 5, 2016 at 17:35

Coach Roebuck

1,0343 gold badges14 silver badges22 bronze badges

Comments

Jake Lin · Accepted Answer · 2018-02-15 05:07:01Z

----- Updated -----

As @ThomasW mentioned, for Swift 4, we need to use String(describing:type(of:self))

----- Old post ----

I prefer to use String(self.dynamicType)

Use it in my project https://github.com/JakeLin/SwiftWeather/blob/e7165b0919dda53fd7fcba9b43fdfe150d73c6f8/SwiftWeather/ForecastView.swift#L135

For Swift 4 this no longer works, but String(describing:type(of:self)) does.
@ThomasW Thanks, have updated the post to reflect the changes.

David Moles · Accepted Answer · 2014-07-17 03:05:25Z

1

If you want to have only the name of the class in swift you can parse the string returned by NSStringFromClass().

This is done in nameOfClass.swift of the INSwift Github repository: https://github.com/indieSoftware/INSwift

edited Jul 17, 2014 at 3:05

David Moles

51.7k30 gold badges152 silver badges249 bronze badges

answered Jun 18, 2014 at 12:14

Sven

111 bronze badge

1 Comment

David Moles Over a year ago

Note that this only works with class types -- if you try it with say a protocol Swift will complain error: type 'MyProtocol' does not conform to protocol 'AnyObject'

TheCodingArt · Accepted Answer · 2015-02-06 22:13:01Z

You shouls now be able to use the following to retrieve the class name in swift

let nameSpaceClassName = NSStringFromClass(RWTCountryPopoverViewController.self) let className = nameSpaceClassName.componentsSeparatedByString(".").last! as String

nacho4d · Accepted Answer · 2015-03-16 02:07:09Z

This is a bit shorter. No need of NSStringFromClass

MyObject.self.description().componentsSeparatedByString(".").last!

Rafat touqir Rafsun · Accepted Answer · 2016-11-10 07:13:57Z

Here is Swift 3+, Xcode 8+ example with code:

class MySuperbClass{ let a = 4 func getClassName() -> String? { return String(describing: type(of:self)).components(separatedBy: ".").first } } let className = String(describing: type(of:MySuperbClass.self)).components(separatedBy: ".").first //className = "MySuperbClass" let classNameFromObject = MySuperbClass().getClassName() //classNameFromObject = "MySuperbClass"

DMagaro · Accepted Answer · 2017-12-26 19:19:25Z

1

Swift 4

super.init(nibName:String(describing:MyClass.self), bundle: nil)

answered Dec 26, 2017 at 19:19

DMagaro

111 bronze badge

Comments

Rolf Hendriks · Accepted Answer · 2014-09-27 03:43:25Z

myObject.description().componentsSeparatedByString(" ").first!

This is not exactly what you want - it will add an unwanted leading '<' and trailing ':'. But when I am debugging I value speed over neatness so this quick + dirty trick worked for me.

anoop4real · Accepted Answer · 2017-04-27 10:06:54Z

0

Swift 3

NSStringFromClass(self).components(separatedBy: ".").last!

answered Apr 27, 2017 at 10:06

anoop4real

7,7184 gold badges56 silver badges59 bronze badges

Comments

Nandish · Accepted Answer · 2020-06-23 08:02:32Z

In latest version of swift, below works for me:

NSStringFromClass(type(of: device!.self))

Collectives™ on Stack Overflow

Get a user-readable version of the class name in swift (in objc NSStringFromClass was fine)

15 Answers 15

3 Comments

3 Comments

Comments

Comments

1 Comment

Comments

2 Comments

1 Comment

Comments

Comments

Comments

Comments

Comments

Comments

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

15 Answers 15

3 Comments

3 Comments

Comments

Comments

1 Comment

Comments

2 Comments

1 Comment

Comments

Comments

Comments

Comments

Comments

Comments

Comments

Linked

Related