Let's say I have a class:
class Fruit {
var fruitName: String
init(getFruit name: String) {
fruitName = name
}
}
Is there any difference between using the constructor, and using .init?
var apple = Fruit(getFruit: "apple")
var orange = Fruit.init(getFruit: "orange")
Im not seeing any difference in playground.
I apologize if the question is badly worded or has been asked before.
From the Initializer Expression section of the language guide:
If you specify a type by name, you can access the type’s initializer without using an initializer expression. In all other cases, you must use an initializer expression.
let s1 = SomeType.init(data: 3) // Valid
let s2 = SomeType(data: 1) // Also valid
let s3 = type(of: someValue).init(data: 7) // Valid
let s4 = type(of: someValue)(data: 5) // Error
Initializing using the explicit .init on the type directly is no different than without it; they are equivalent from Swift's perspective, so most folks prefer the brevity of omitting .init.
To my knowledge, there is absolutely no difference.
It is generally convention in Swift to call the constructor without the .init This is the 'swiftier' shorthand.
I was trying to do something like this (it is a contrived example for demonstration purposes only):
class Test {
let hello = "hello"
let world = "world"
let phrase: String {
return self.hello + self.world
}
}
but you can't use let for computed properties in Swift. Is there a way to do this without having to write an init() method? Thanks!
The reason let doesn't work on a read-only calculated property is because it's used to state that the property's actual value will never change after being set – not that the property is read-only. As the Apple docs say (emphasis mine):
You must declare computed properties — including read-only computed
properties — as variable properties with the var keyword, because their
value is not fixed. The let keyword is only used for constant
properties, to indicate that their values cannot be changed once they
are set as part of instance initialization.
You therefore need to use var in order to reflect the fact that a calculated property's value could change at any time, as you're creating it on the fly when accessing it. Although in your code, this can't happen – as your hello and world properties are let constants themselves. However, Swift is unable to infer this, so you still have to use var.
For example:
class Test {
let hello = "hello"
let world = "world"
var phrase: String {
return self.hello + self.world
}
}
(This doesn't change the readability of the property – as because you haven't provided it with a setter, it's still read-only)
However in your case, you might want to consider using a lazy property instead, as your hello and world properties are constants. A lazy property is created when it's first accessed, and keeps its value for the rest of its lifetime – meaning you won't have to keep on concatenating two constants together every time you access it.
For example:
class Test {
let hello = "hello"
let world = "world"
lazy var phrase: String = {
return self.hello + self.world
}()
}
Another characteristic of let properties is that their value should always be known before initialisation. Because the value of a lazy property might not be known before then, you also need to define it as a var.
If you're still adamant on wanting a let property for this, then as far as I can see, you have two options.
The first is the neatest (although you've said you don't want to do it) – you can assign your phrase property in the initialiser. As long as you do this before the super.init call, you don't have to deal with optionals. For example:
class Test {
let hello = "hello"
let world = "world"
let phrase: String
init() {
phrase = hello+world
}
}
You simply cannot do it inline, as self at that scope refers to the static class, not an instance of the class. Therefore you cannot access the instance members, and have to use init() or a lazy/calculated property.
The second option is pretty hacky – you can mirror your hello and world properties at class level, so you can therefore access them inline in your phrase declaration. For example:
class Test {
static let hello = "hello"
static let world = "world"
// for some reason, Swift has trouble inferring the type
// of the static mirrored versions of these properties
let hello:String = Test.hello
let world:String = Test.world
let phrase = hello+world
}
If you don't actually need your hello or world properties as instance properties, then you can just make them static – which will solve your problem.
Yes to make it work as computed properties, replace let to var.
Like,
class Test {
let hello = "hello"
let world = "world"
var phrase: String {
return self.hello + self.world
}
}
This way you can use it without init()
Is there a way to get the compile time name of a variable in Swift 2?
I mean the first variable name, which references to a new class instance, if any.
Here is a simple example:
public class Parameter : FloatLiteralConvertible {
var name:String?
var value:Double
// init from float literal
public required init (floatLiteral value: FloatLiteralType) {
self.value = Double(value)
self.name = getLiteralName()
}
func getLiteralName () -> String {
var literalName:String = ""
// do some magic to return the name
return literalName
}
}
let x:Parameter = 2.0
print(x.value) // this returns "2.0"
print(x.name!) // I want this to return "x"
I've already checked similar questions on that topic handling mirroring or objective-c reflections. But in all those cases, one can get only the property names in a class - in the example above name and value.
The same question has been asked in 2014 - Swift: Get Variable Actual Name as String
- and I hope, that since then there is a solution in swift 2.
No, there is no way to do that.
You have to understand that in the compiled state that variable usually does not exist. It can be optimized out or it is represented only as an item on the execution stack.
Even in languages with much better reflection that Swift has, usually you cannot inspect local variables.
To be honest, getting the name of a local variable dynamically has no practical use case.
Is there a way to print the runtime type of a variable in swift? For example:
var now = NSDate()
var soon = now.dateByAddingTimeInterval(5.0)
println("\(now.dynamicType)")
// Prints "(Metatype)"
println("\(now.dynamicType.description()")
// Prints "__NSDate" since objective-c Class objects have a "description" selector
println("\(soon.dynamicType.description()")
// Compile-time error since ImplicitlyUnwrappedOptional<NSDate> has no "description" method
In the example above, I'm looking for a way to show that the variable "soon" is of type ImplicitlyUnwrappedOptional<NSDate>, or at least NSDate!.
Update September 2016
Swift 3.0: Use type(of:), e.g. type(of: someThing) (since the dynamicType keyword has been removed)
Update October 2015:
I updated the examples below to the new Swift 2.0 syntax (e.g. println was replaced with print, toString() is now String()).
From the Xcode 6.3 release notes:
#nschum points out in the comments that the Xcode 6.3 release notes show another way:
Type values now print as the full demangled type name when used with
println or string interpolation.
import Foundation
class PureSwiftClass { }
var myvar0 = NSString() // Objective-C class
var myvar1 = PureSwiftClass()
var myvar2 = 42
var myvar3 = "Hans"
print( "String(myvar0.dynamicType) -> \(myvar0.dynamicType)")
print( "String(myvar1.dynamicType) -> \(myvar1.dynamicType)")
print( "String(myvar2.dynamicType) -> \(myvar2.dynamicType)")
print( "String(myvar3.dynamicType) -> \(myvar3.dynamicType)")
print( "String(Int.self) -> \(Int.self)")
print( "String((Int?).self -> \((Int?).self)")
print( "String(NSString.self) -> \(NSString.self)")
print( "String(Array<String>.self) -> \(Array<String>.self)")
Which outputs:
String(myvar0.dynamicType) -> __NSCFConstantString
String(myvar1.dynamicType) -> PureSwiftClass
String(myvar2.dynamicType) -> Int
String(myvar3.dynamicType) -> String
String(Int.self) -> Int
String((Int?).self -> Optional<Int>
String(NSString.self) -> NSString
String(Array<String>.self) -> Array<String>
Update for Xcode 6.3:
You can use the _stdlib_getDemangledTypeName():
print( "TypeName0 = \(_stdlib_getDemangledTypeName(myvar0))")
print( "TypeName1 = \(_stdlib_getDemangledTypeName(myvar1))")
print( "TypeName2 = \(_stdlib_getDemangledTypeName(myvar2))")
print( "TypeName3 = \(_stdlib_getDemangledTypeName(myvar3))")
and get this as output:
TypeName0 = NSString
TypeName1 = __lldb_expr_26.PureSwiftClass
TypeName2 = Swift.Int
TypeName3 = Swift.String
Original answer:
Prior to Xcode 6.3 _stdlib_getTypeName got the mangled type name of a variable. Ewan Swick's blog entry helps to decipher these strings:
e.g. _TtSi stands for Swift's internal Int type.
Mike Ash has a great blog entry covering the same topic.
Edit: A new toString function has been introduced in Swift 1.2 (Xcode 6.3).
You can now print the demangled type of any type using .self and any instance using .dynamicType:
struct Box<T> {}
toString("foo".dynamicType) // Swift.String
toString([1, 23, 456].dynamicType) // Swift.Array<Swift.Int>
toString((7 as NSNumber).dynamicType) // __NSCFNumber
toString((Bool?).self) // Swift.Optional<Swift.Bool>
toString(Box<SinkOf<Character>>.self) // __lldb_expr_1.Box<Swift.SinkOf<Swift.Character>>
toString(NSStream.self) // NSStream
Try calling YourClass.self and yourObject.dynamicType.
Reference: https://devforums.apple.com/thread/227425.
Swift 3.0
let string = "Hello"
let stringArray = ["one", "two"]
let dictionary = ["key": 2]
print(type(of: string)) // "String"
// Get type name as a string
String(describing: type(of: string)) // "String"
String(describing: type(of: stringArray)) // "Array<String>"
String(describing: type(of: dictionary)) // "Dictionary<String, Int>"
// Get full type as a string
String(reflecting: type(of: string)) // "Swift.String"
String(reflecting: type(of: stringArray)) // "Swift.Array<Swift.String>"
String(reflecting: type(of: dictionary)) // "Swift.Dictionary<Swift.String, Swift.Int>"
Is this what you're looking for?
println("\(object_getClassName(now))");
It prints "__NSDate"
UPDATE: Please note this no longer seems to work as of Beta05
My current Xcode is Version 6.0 (6A280e).
import Foundation
class Person { var name: String; init(name: String) { self.name = name }}
class Patient: Person {}
class Doctor: Person {}
var variables:[Any] = [
5,
7.5,
true,
"maple",
Person(name:"Sarah"),
Patient(name:"Pat"),
Doctor(name:"Sandy")
]
for variable in variables {
let typeLongName = _stdlib_getDemangledTypeName(variable)
let tokens = split(typeLongName, { $0 == "." })
if let typeName = tokens.last {
println("Variable \(variable) is of Type \(typeName).")
}
}
Output:
Variable 5 is of Type Int.
Variable 7.5 is of Type Double.
Variable true is of Type Bool.
Variable maple is of Type String.
Variable Swift001.Person is of Type Person.
Variable Swift001.Patient is of Type Patient.
Variable Swift001.Doctor is of Type Doctor.
As of Xcode 6.3 with Swift 1.2, you can simply convert type values into the full demangled String.
toString(Int) // "Swift.Int"
toString(Int.Type) // "Swift.Int.Type"
toString((10).dynamicType) // "Swift.Int"
println(Bool.self) // "Swift.Bool"
println([UTF8].self) // "Swift.Array<Swift.UTF8>"
println((Int, String).self) // "(Swift.Int, Swift.String)"
println((String?()).dynamicType)// "Swift.Optional<Swift.String>"
println(NSDate) // "NSDate"
println(NSDate.Type) // "NSDate.Type"
println(WKWebView) // "WKWebView"
toString(MyClass) // "[Module Name].MyClass"
toString(MyClass().dynamicType) // "[Module Name].MyClass"
You can still access the class, through className (which returns a String).
There are actually several ways to get the class, for example classForArchiver, classForCoder, classForKeyedArchiver (all return AnyClass!).
You can't get the type of a primitive (a primitive is not a class).
Example:
var ivar = [:]
ivar.className // __NSDictionaryI
var i = 1
i.className // error: 'Int' does not have a member named 'className'
If you want to get the type of a primitive, you have to use bridgeToObjectiveC(). Example:
var i = 1
i.bridgeToObjectiveC().className // __NSCFNumber
You can use reflect to get information about object.
For example name of object class:
var classname = reflect(now).summary
Xcode 8 Swift 3.0 use type(of:)
let className = "\(type(of: instance))"
I had luck with:
let className = NSStringFromClass(obj.dynamicType)
SWIFT 5
With the latest release of Swift 3 we can get pretty descriptions of type names through the String initializer. Like, for example print(String(describing: type(of: object))). Where object can be an instance variable like array, a dictionary, an Int, a NSDate, an instance of a custom class, etc.
Here is my complete answer: Get class name of object as string in Swift
That question is looking for a way to getting the class name of an object as string but, also i proposed another way to getting the class name of a variable that isn't subclass of NSObject. Here it is:
class Utility{
class func classNameAsString(obj: Any) -> String {
//prints more readable results for dictionaries, arrays, Int, etc
return String(describing: type(of: obj))
}
}
I made a static function which takes as parameter an object of type Any and returns its class name as String :) .
I tested this function with some variables like:
let diccionary: [String: CGFloat] = [:]
let array: [Int] = []
let numInt = 9
let numFloat: CGFloat = 3.0
let numDouble: Double = 1.0
let classOne = ClassOne()
let classTwo: ClassTwo? = ClassTwo()
let now = NSDate()
let lbl = UILabel()
and the output was:
diccionary is of type Dictionary
array is of type Array
numInt is of type Int
numFloat is of type CGFloat
numDouble is of type Double
classOne is of type: ClassOne
classTwo is of type: ClassTwo
now is of type: Date
lbl is of type: UILabel
In Xcode 8, Swift 3.0
Steps:
1. Get the Type:
Option 1:
let type : Type = MyClass.self //Determines Type from Class
Option 2:
let type : Type = type(of:self) //Determines Type from self
2. Convert Type to String:
let string : String = "\(type)" //String
In Swift 3.0, you can use type(of:), as dynamicType keyword has been removed.
To get a type of object or class of object in Swift, you must need to use a type(of: yourObject)
type(of: yourObject)
When using Cocoa (not CocoaTouch), you can use the className property for objects that are subclasses of NSObject.
println(now.className)
This property is not available for normal Swift objects, which aren't subclasses of NSObject (and in fact, there is no root id or object type in Swift).
class Person {
var name: String?
}
var p = Person()
println(person.className) // <- Compiler error
In CocoaTouch, at this time there is not a way to get a string description of the type of a given variable. Similar functionality also does not exist for primitive types in either Cocoa or CocoaTouch.
The Swift REPL is able to print out a summary of values including its type, so it is possible this manner of introspection will be possible via an API in the future.
EDIT: dump(object) seems to do the trick.
The top answer doesn't have a working example of the new way of doing this using type(of:. So to help rookies like me, here is a working example, taken mostly from Apple's docs here - https://developer.apple.com/documentation/swift/2885064-type
doubleNum = 30.1
func printInfo(_ value: Any) {
let varType = type(of: value)
print("'\(value)' of type '\(varType)'")
}
printInfo(doubleNum)
//'30.1' of type 'Double'
I've tried some of the other answers here but milage seems to very on what the underling object is.
However I did found a way you can get the Object-C class name for an object by doing the following:
now?.superclass as AnyObject! //replace now with the object you are trying to get the class name for
Here is and example of how you would use it:
let now = NSDate()
println("what is this = \(now?.superclass as AnyObject!)")
In this case it will print NSDate in the console.
I found this solution which hopefully might work for someone else.
I created a class method to access the value. Please bear in mind this will work for NSObject subclass only. But at least is a clean and tidy solution.
class var className: String!{
let classString : String = NSStringFromClass(self.classForCoder())
return classString.componentsSeparatedByString(".").last;
}
In the latest XCode 6.3 with Swift 1.2, this is the only way I found:
if view.classForCoder.description() == "UISegment" {
...
}
Many of the answers here do not work with the latest Swift (Xcode 7.1.1 at time of writing).
The current way of getting the information is to create a Mirror and interrogate that. For the classname it is as simple as:
let mirror = Mirror(reflecting: instanceToInspect)
let classname:String = mirror.description
Additional information about the object can also be retrieved from the Mirror. See http://swiftdoc.org/v2.1/type/Mirror/ for details.
Swift version 4:
print("\(type(of: self)) ,\(#function)")
// within a function of a class
Thanks #Joshua Dance
In lldb as of beta 5, you can see the class of an object with the command:
fr v -d r shipDate
which outputs something like:
(DBSalesOrderShipDate_DBSalesOrderShipDate_ *) shipDate = 0x7f859940
The command expanded out means something like:
Frame Variable (print a frame variable) -d run_target (expand dynamic types)
Something useful to know is that using "Frame Variable" to output variable values guarantees no code is executed.
I've found a solution for self-developed classes (or such you have access to).
Place the following computed property within your objects class definition:
var className: String? {
return __FILE__.lastPathComponent.stringByDeletingPathExtension
}
Now you can simply call the class name on your object like so:
myObject.className
Please note that this will only work if your class definition is made within a file that is named exactly like the class you want the name of.
As this is commonly the case the above answer should do it for most cases. But in some special cases you might need to figure out a different solution.
If you need the class name within the class (file) itself you can simply use this line:
let className = __FILE__.lastPathComponent.stringByDeletingPathExtension
Maybe this method helps some people out there.
Based on the answers and comments given by Klass and Kevin Ballard above, I would go with:
println(_stdlib_getDemangledTypeName(now).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(soon).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(soon?).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(soon!).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(myvar0).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(myvar1).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(myvar2).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(myvar3).componentsSeparatedByString(".").last!)
which will print out:
"NSDate"
"ImplicitlyUnwrappedOptional"
"Optional"
"NSDate"
"NSString"
"PureSwiftClass"
"Int"
"Double"
let i: Int = 20
func getTypeName(v: Any) -> String {
let fullName = _stdlib_demangleName(_stdlib_getTypeName(i))
if let range = fullName.rangeOfString(".") {
return fullName.substringFromIndex(range.endIndex)
}
return fullName
}
println("Var type is \(getTypeName(i)) = \(i)")
Swift 4:
// "TypeName"
func stringType(of some: Any) -> String {
let string = (some is Any.Type) ? String(describing: some) : String(describing: type(of: some))
return string
}
// "ModuleName.TypeName"
func fullStringType(of some: Any) -> String {
let string = (some is Any.Type) ? String(reflecting: some) : String(reflecting: type(of: some))
return string
}
Usage:
print(stringType(of: SomeClass())) // "SomeClass"
print(stringType(of: SomeClass.self)) // "SomeClass"
print(stringType(of: String())) // "String"
print(fullStringType(of: String())) // "Swift.String"
There appears to be no generic way to print the type name of an arbitrary value's type. As others have noted, for class instances you can print value.className but for primitive values it appears that at runtime, the type information is gone.
For instance, it looks as if there's not a way to type: 1.something() and get out Int for any value of something. (You can, as another answer suggested, use i.bridgeToObjectiveC().className to give you a hint, but __NSCFNumber is not actually the type of i -- just what it will be converted to when it crosses the boundary of an Objective-C function call.)
I would be happy to be proven wrong, but it looks like the type checking is all done at compile time, and like C++ (with RTTI disabled) much of the type information is gone at runtime.
This is how you get a type string of your object or Type which is consistent and takes into account to which module the object definition belongs to or nested in. Works in Swift 4.x.
#inline(__always) func typeString(for _type: Any.Type) -> String {
return String(reflecting: type(of: _type))
}
#inline(__always) func typeString(for object: Any) -> String {
return String(reflecting: type(of: type(of: object)))
}
struct Lol {
struct Kek {}
}
// if you run this in playground the results will be something like
typeString(for: Lol.self) // __lldb_expr_74.Lol.Type
typeString(for: Lol()) // __lldb_expr_74.Lol.Type
typeString(for: Lol.Kek.self)// __lldb_expr_74.Lol.Kek.Type
typeString(for: Lol.Kek()) // __lldb_expr_74.Lol.Kek.Type
Not exactly what you are after, but you can also check the type of the variable against Swift types like so:
let object: AnyObject = 1
if object is Int {
}
else if object is String {
}
For example.
Xcode 7.3.1, Swift 2.2:
String(instanceToPrint.self).componentsSeparatedByString(".").last