Answering The Most Common Questions About Swift by@Kiran

Answering The Most Common Questions About Swift

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Content Writer at Truemark Technology. Company Website Link -

Swift is a general-purpose programming language developed by Apple Inc. It is a powerful language developed for iOS, iPadOS, macOS, watchOS, tvOS, Linux, and z/OS. Swift is one of the popular languages and has a huge scope if you want to develop apps for Apple products. So, today we will be checking out the 11 most asked Swift programming questions.

11 Most Asked Swift Programming Questions

1. How to call Objective-C code from Swift?


Using Objective-C Classes in Swift 

If you have an existing class that you'd like to use, perform Step 2 and then skip to Step 5. (For some cases, add an explicit

#import <Foundation/Foundation.h to an older Objective-C File.) 

Step 1: Add Objective-C Implementation -- .m 

Add a

file to your class, and name it

Step 2: Add Bridging Header 

When adding your

file, you'll likely be hit with a prompt that looks like this:


Click Yes

If you did not see the prompt or accidentally deleted your bridging header, add a new

file to your project and name it <

In some situations, particularly when working with Objective-C frameworks, you don't add an Objective-C class explicitly and Xcode can't find the linker. In this case, create your

file named as mentioned above, then make sure you link its path in your target's project settings like so: 



It's best practice to link your project using the

macro so that if you move your project, or work on it with others using a remote repository, it will still work.
can be thought of as the directory that contains your .xcodeproj file. It might look like this:


Step 3: Add Objective-C Header -- .h

Add another

file and name it

Step 4: Build your Objective-C Class



#import <Foundation/Foundation.h>

@interface CustomObject : NSObject

@property (strong, nonatomic) id someProperty;

- (void) someMethod;




#import "CustomObject.h"

@implementation CustomObject 

- (void) someMethod {
    NSLog(@"SomeMethod Ran");}


Step 5: Add Class to Bridging-Header



#import "CustomObject.h"

Step 6: Use your Object



var instanceOfCustomObject = CustomObject()
instanceOfCustomObject.someProperty = "Hello World"

There is no need to import explicitly; that's what the bridging header is for.

Using Swift Classes in Objective-C 

Step 1: Create New Swift Class 

Add a

file to your project, and name it



import Foundation

@objc(MySwiftObject)class MySwiftObject : NSObject {

    var someProperty: AnyObject = "Some Initializer Val" as NSString

    init() {}

    func someFunction(someArg: Any) -> NSString {
        return "You sent me \(someArg)"

Step 2: Import Swift Files to ObjC Class



#import "<#YourProjectName#>-Swift.h"

The file:

should already be created automatically in your project, even if you can not see it.

Step 3: Use your class

MySwiftObject * myOb = [MySwiftObject new];
NSLog(@"MyOb.someProperty: %@", myOb.someProperty);
myOb.someProperty = @"Hello World";
NSLog(@"MyOb.someProperty: %@", myOb.someProperty);

NSString * retString = [myOb someFunctionWithSomeArg:@"Arg"];

NSLog(@"RetString: %@", retString);


  1. If Code Completion isn't behaving as you expect, try running a quick build with
    to help Xcode find some of the Objective-C code from a Swift context and vice versa.
  2. If you add a
    file to an older project and get the error
    dyld: Library not loaded: @rpath/libswift_stdlib_core.dylib
    , try completely restarting Xcode.
  3. While it was originally possible to use pure Swift classes (Not descendents of
    ) which are visible to Objective-C by using the
    prefix, this is no longer possible. Now, to be visible in Objective-C, the Swift object must either be a class conforming to
    (easiest way to do this is to inherit from
    ), or to be an
    with a raw value of some integer type like

2. Is there a stand-in for #pragma mark in Swift?


You can use

 // MARK

There has also been discussion that liberal use of class extensions might be a better practice anyway. Since extensions can implement protocols, you can e.g. put all of your table view delegate methods in an extension and group your code at a more semantic level than

#pragma mark
is capable of. 

Alternative Answer: 

Up to Xcode 5 the preprocessor directive

#pragma mark

From Xcode 6 on, you have to use


These preprocessor features allow to bring some structure to the function drop down box of the source code editor.

Some examples :

// MARK:

-> will be preceded by a horizontal divider

// MARK: your text goes here

-> puts 'your text goes here' in bold in the drop down list

// MARK: - your text goes here

-> puts 'your text goes here' in bold in the drop down list, preceded by a horizontal divider

Added screenshot 'cause some people still seem to have issues with this :


3. How to get the length of a


As of Swift 4

It’s just:


As of Swift 2:

With Swift 2, Apple has changed global functions to protocol extensions, extensions that match any type conforming to a protocol. Thus the new syntax is:


As of Swift 1

Use the count characters method:

let unusualMenagerie = "Koala &#128040;, Snail &#128012;, Penguin &#128039;, Dromedary &#128042;"
println("unusualMenagerie has \(count(unusualMenagerie)) characters")
// prints "unusualMenagerie has 40 characters"

right from the Apple Swift Guide

(Note: For versions of Swift earlier than 1.2, this would be

instead) for your variable, it would be

length = count(test1) // was countElements in earlier versions of Swift

Or you can use test1.utf16count

Alternative Answer:

For Swift 2.0 and 3.0, use

. But, there are a few things you should know. So, read on.

Counting characters in Swift

Before Swift 2.0, count was a global function. As of Swift 2.0, it can be called as a member function.


It will return the actual number of Unicode characters in a

, so it’s the most correct alternative in the sense that, if you’d print the string and count characters by hand, you’d get the same result.

However, because of the way

are implemented in Swift, characters don’t always take up the same amount of memory, so be aware that this behaves quite differently than the usual character count methods in other languages.

For example, you can also use


But, as noted below, the returned value is not guaranteed to be the same as that of calling


From the language reference:

Extended grapheme clusters can be composed of one or more Unicode scalars. This means that different characters—and different representations of the same character—can require different amounts of memory to store. Because of this, characters in Swift do not each take up the same amount of memory within a string’s representation. As a result, the number of characters in a string cannot be calculated without iterating through the string to determine its extended grapheme cluster boundaries. If you are working with particularly long string values, be aware that the characters property must iterate over the Unicode scalars in the entire string in order to determine the characters for that string.
The count of the characters returned by the characters property is not always the same as the length property of an NSString that contains the same characters. The length of an NSString is based on the number of 16-bit code units within the string’s UTF-16 representation and not the number of Unicode extended grapheme clusters within the string.

An example that perfectly illustrates the situation described above
is that of checking the length of a string containing a single emoji

var emoji = "👍"
emoji.characters.count             //returns 1
emoji.utf16.count                  //returns 2

4. How to iterate a loop with index and element in Swift?


As of Swift 3.0, if you need the index for each element along with its value, you can use the

method to iterate over the array. It returns a sequence of pairs composed of the index and the value for each item in the array. For example:

for (index, element) in list.enumerated() {
  print("Item \(index): \(element)")}

Before Swift 3.0 and after Swift 2.0, the function was called


for (index, element) in list.enumerate() {
    print("Item \(index): \(element)")}

Prior to Swift 2.0,

was a global function.

for (index, element) in enumerate(list) {
    println("Item \(index): \(element)")}

Alternative Answer:

Swift 5 provides a method called 

for Array.
has the following declaration:

func enumerated() -> EnumeratedSequence<Array<Element>>
Returns a sequence of pairs (n, x), where n represents a consecutive integer starting at zero and x represents an element of the sequence.

In the simplest cases, you may use 

) with a for loop. For example:

let list = ["Car", "Bike", "Plane", "Boat"]
for (index, element) in list.enumerated() {
    print(index, ":", element)

0 : Car
1 : Bike
2 : Plane
3 : Boat

Note: However that you’re not limited to use 

 with a for loop. In fact, if you plan to use 
 with a for loop for something similar to the following code, you’re doing it wrong:

let list = [Int](1...5)
var arrayOfTuples = [(Int, Int)]()

for (index, element) in list.enumerated() {
    arrayOfTuples += [(index, element)]}

print(arrayOfTuples) // prints [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]

A swiftier way to do this is:

let list = [Int](1...5)
let arrayOfTuples = Array(list.enumerated())
print(arrayOfTuples) // prints [(offset: 0, element: 1), (offset: 1, element: 2), (offset: 2, element: 3), (offset: 3, element: 4), (offset: 4, element: 5)]

As an alternative, you may also use 


let list = [Int](1...5)
let arrayOfDictionaries = list.enumerated().map { (a, b) in return [a : b] }
print(arrayOfDictionaries) // prints [[0: 1], [1: 2], [2: 3], [3: 4], [4: 5]]

Moreover, although it has some limitations

 can be a good replacement to a for loop:

let list = [Int](1...5)
list.reversed().enumerated().forEach { print($0, ":", $1) }

0 : 5
1 : 4
2 : 3
3 : 2
4 : 1

By using 

, you can even iterate manually on your 
. For example:

import UIKitimport PlaygroundSupport

class ViewController: UIViewController {

    var generator = ["Car", "Bike", "Plane", "Boat"].enumerated().makeIterator()

    override func viewDidLoad() {

        let button = UIButton(type: .system)
        button.setTitle("Tap", for: .normal)
        button.frame = CGRect(x: 100, y: 100, width: 100, height: 100)
        button.addTarget(self, action: #selector(iterate(_:)), for: .touchUpInside)

    @objc func iterate(_ sender: UIButton) {
        let tuple =
        print(String(describing: tuple))


PlaygroundPage.current.liveView = ViewController()

 Optional((offset: 0, element: "Car"))
 Optional((offset: 1, element: "Bike"))
 Optional((offset: 2, element: "Plane"))
 Optional((offset: 3, element: "Boat"))

5. Is there a replacement/solution for #ifdef where you can define a macro using compiler preprocessors?


In Swift, you can still use the "#if/#else/#endif" preprocessor macros (although more constrained), as per Apple docs. Here's an example:

    let a = 2
    let a = 3

Now, you must set the "DEBUG" symbol elsewhere, though. Set it in the "Swift Compiler - Custom Flags" section, "Other Swift Flags" line. You add the DEBUG symbol with the


As usual, you can set a different value when in Debug or when in Release.

It works, but it doesn't seem to be recognized in a playground though.

Important Note:

doesn't work. Only
works. Seems compiler is ignoring a flag with a specific value.

Alternative Answer:

As stated in Apple Docs

The Swift compiler does not include a preprocessor. Instead, it takes advantage of compile-time attributes, build configurations, and language features to accomplish the same functionality. For this reason, preprocessor directives are not imported in Swift.

By using custom Build Configurations:

  1. Go to your project / select your target / Build Settings / search for Custom Flags
  2. For your chosen target set your custom flag using -D prefix (without white spaces), for both Debug and Release
  3. Do above steps for every target you have

Here's how you check for target:

    print("We have a banana")
#elseif MELONA

Tested using Swift 2.2

6. How to split a string into an array in Swift?


The Swift way is to use the global split function, like so:

var fullName = "First Last"
var fullNameArr = split(fullName) {$0 == " "}
var firstName: String = fullNameArr[0]
var lastName: String? = fullNameArr.count > 1 ? fullNameArr[1] : nilW

With Swift 2

In Swift 2 the use of split becomes a bit more complicated due to the introduction of the internal CharacterView type. This means that String no longer adopts the SequenceType or CollectionType protocols and you must instead use the

property to access a CharacterView type representation of a String instance. (Note: CharacterView does adopt SequenceType and CollectionType protocols).

let fullName = "First Last"
let fullNameArr = fullName.characters.split{$0 == " "}.map(String.init)// or simply:
// let fullNameArr = fullName.characters.split{" "}.map(String.init)

fullNameArr[0] // First
fullNameArr[1] // Last 

Alternative Answer:

Just call


import Foundation

var fullName: String = "First Last"let fullNameArr = fullName.componentsSeparatedByString(" ")

var firstName: String = fullNameArr[0]
var lastName: String = fullNameArr[1]

Update for Swift 3+

import Foundation

let fullName    = "First Last"
let fullNameArr = fullName.components(separatedBy: " ")

let name    = fullNameArr[0]
let surname = fullNameArr[1]

7. How to use @selector() in Swift?


Swift itself doesn't use selectors — several design patterns that in Objective-C make use of selectors work differently in Swift. (For example, use optional chaining on protocol types or

tests instead of
, and use closures wherever you can instead of performSelector: for better type/memory safety.) 

But there are still a number of important ObjC-based APIs that use selectors, including timers and the target/action pattern. Swift provides the

type for working with these. (Swift automatically uses this in place of ObjC's

In Swift 2.2 (Xcode 7.3) and later (including Swift 3 / Xcode 8 and Swift 4 / Xcode 9):

You can construct a

from a Swift function type using the

let timer = Timer(timeInterval: 1, target: object,
                  selector: #selector(MyClass.test),
                  userInfo: nil, repeats: false)
button.addTarget(object, action: #selector(MyClass.buttonTapped),
                 for: .touchUpInside)
             with: button, with: otherButton)

The great thing about this approach? A function reference is checked by the Swift compiler, so you can use the

expression only with class/method pairs that actually exist and are eligible for use as selectors (see "Selector availability" below). You're also free to make your function reference only as specific as you need, as per the Swift 2.2+ rules for function-type naming.

(This is actually an improvement over ObjC's

directive, because the compiler's -
check verifies only that the named selector exists. The Swift function reference you pass to
checks existence, membership in a class, and type signature.) There are a couple of extra caveats for the function references you pass to the

  • Multiple functions with the same base name can be differentiated by their parameter labels using the aforementioned syntax for function references (e.g.
    . But if a function has no parameters, the only way to disambiguate it is to use an
    cast with the function's type signature (e.g.
    foo as () -> () vs foo(_:))
  • There's a special syntax for property getter/setter pairs in Swift 3.0+. For example, given a
    var foo: Int
    , you can use

General notes:

Cases where

doesn't work, and naming: Sometimes you don't have a function reference to make a selector with (for example, with methods dynamically registered in the ObjC runtime). In that case, you can construct a
from a string: e.g.
— though you lose the compiler's validity checking. When you do that, you need to follow ObjC naming rules, including colons (:) for each parameter. 

Selector availability: The method referenced by the selector must be exposed to the ObjC runtime. In Swift 4, every method exposed to ObjC must have its declaration prefaced with the

attribute. (In previous versions you got that attribute for free in some cases, but now you have to explicitly declare it.) 

Remember that

symbols aren't exposed to the runtime, too — your method needs to have at least

Key paths: These are related to but not quite the same as selectors. There's a special syntax for these in Swift 3, too: e.g.

chris.valueForKeyPath #keyPath(Person.friends.firstName))
. See SE-0062 for details. And even more KeyPath stuff in Swift 4, so make sure you're using the right KeyPath-based API instead of selectors if appropriate. 

You can read more about selectors under Interacting with Objective-C APIs in Using Swift with Cocoa and Objective-C

Note: Before Swift 2.2,

conformed to
, so you might find old code where bare strings are passed to APIs that take selectors. You'll want to run "Convert to Current Swift Syntax" in Xcode to get those using

Alternative Answer:

Here's a quick example of how to use the

class on Swift:

override func viewDidLoad() {

    var rightButton = UIBarButtonItem(title: "Title", style: UIBarButtonItemStyle.Plain, target: self, action: Selector("method"))
    self.navigationItem.rightBarButtonItem = rightButton

func method() {
    // Something cool here   

Note that if the method passed as a string doesn't work, it will fail at runtime, not compile time, and crash your app. Be careful.

8. Do Swift-based applications work on OS X 10.9/iOS 7 and lower?


Swift applications compile into standard binaries and can be run on OS X 10.9 and iOS 7. A simple Swift application used for testing:

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: NSDictionary?) -> Bool {
    self.window = UIWindow(frame: UIScreen.mainScreen().bounds)

    var controller = UIViewController()
    var view = UIView(frame: CGRectMake(0, 0, 320, 568))
    view.backgroundColor = UIColor.redColor()
    controller.view = view

    var label = UILabel(frame: CGRectMake(0, 0, 200, 21)) = CGPointMake(160, 284)
    label.textAlignment = NSTextAlignment.Center
    label.text = "I'am a test label"

    self.window!.rootViewController = controller
    return true}

9. What is the proper usage of
singleton model in Swift?


Use the class constant approach if you are using Swift 1.2 or above and the nested struct approach if you need to support earlier versions.

With Swift, there are three approaches to implement the Singleton pattern that support lazy initialization and thread-safety. 

Class constant

class Singleton  {
   static let sharedInstance = Singleton()

This approach supports lazy initialization because Swift lazily initializes class constants (and variables), and is thread safe by the definition of

. This is now officially recommended way to instantiate a singleton.

Class constants were introduced in Swift 1.2. If you need to support an earlier version of Swift, use the nested struct approach below or a global constant.

Nested struct

class Singleton {
    class var sharedInstance: Singleton {
        struct Static {
            static let instance: Singleton = Singleton()
        return Static.instance

Here we are using the static constant of a nested struct as a class constant. This is a workaround for the lack of static class constants in Swift 1.1 and earlier and still works as a workaround for the lack of static constants and variables in functions.


The traditional Objective-C approach ported to Swift.

class Singleton {
    class var sharedInstance: Singleton {
        struct Static {
            static var onceToken: dispatch_once_t = 0
            static var instance: Singleton? = nil
        dispatch_once(&Static.onceToken) {
            Static.instance = Singleton()
        return Static.instance!

See this GitHub project for unit tests.

10. How to make a weak protocol reference in 'pure' Swift (without @objc)?


You need to declare the type of the protocol as


protocol ProtocolNameDelegate: AnyObject {
    // Protocol stuff goes here

class SomeClass {
    weak var delegate: ProtocolNameDelegate?


you say that only classes can conform to this protocol, whereas structs or enums can't.

Supplemental Answer:

  • The purpose of using the
    keyword is to avoid strong reference cycles (retain cycles). Strong reference cycles happen when two class instances have strong references to each other. Their reference counts never go to zero so they never get deallocated.
  • You only need to use
    if the delegate is a class. Swift structs and enums are value types (their values are copied when a new instance is made), not reference types, so they don't make strong reference cycles.
  • weak
    references are always optional (otherwise you would use
    ) and always use
    ) so that the optional can be set to
    when it is deallocated.
  • A parent class should naturally have a strong reference to its child classes and thus not use the
    keyword. When a child wants a reference to its parent, though, it should make it a weak reference by using the
  • weak
    should be used when you want a reference to a class that you don't own, not just for a child referencing its parent. When two non-hierarchical classes need to reference each other, choose one to be weak. The one you choose depends on the situation. See the answers to this question for more on this.
  • As a general rule, delegates should be marked as
    because most delegates are referencing classes that they do not own. This is definitely true when a child is using a delegate to communicate with a parent. Using a weak reference for the delegate is what the documentation recommends. (But see this, too.)
  • Protocols can be used for both reference types (classes) and value types (structs, enums). So in the likely case that you need to make a delegate weak, you have to make it an object-only protocol. The way to do that is to add
    to the protocol's inheritance list. (In the past you did this using the
    keyword, but
    is preferred now
protocol MyClassDelegate: AnyObject {
    // ...

class SomeClass {
    weak var delegate: MyClassDelegate?

11. How to convert Int to String in Swift?




let x : Int = 42
var myString = String(x)

And the other way around - converting


let myString : String = "42"
let x: Int? = myString.toInt()

if (x != nil) {
    // Successfully converted String to Int

Or if you're using Swift 2 or 3:

let x: Int? = Int(myString)

Alternative Answer:

Check the Below Answer:

let x : Int = 45
var stringValue = "\(x)"

In Conclusion 

These are the 11 most commonly asked Swift programming questions. If you have any suggestions or any confusion, please comment below. If you need any help, we will be glad to help you. 

Hope this article helped you.

This post was first published on DevPost by Truemark.


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