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var fruits = ["apples", "oranges", "PAPAYA", "kiwi"]fruits.sort { (a: String, b: String) -> Bool in return a < b}print(fruits)

There are three ways in which we can simplify the given code snippet:

Apple gadgets are used by BILLIONS of people all over the world. The number of iOS users has been constantly INCREASING, which is good news for iOS application developers. iOS Swift developers MUST also keep up with the latest developments in the iOS Swift community. Make sure you are keeping up with Apple developer news, podcasts, and blogs. This is unlikely to be a question in an interview, but it distinguishes you. We hope that the Swift Interview questions answered here are extremely helpful in studying the fundamentals and advanced topics of iOS Swift. Any beginner or experienced professional who understands these Swift and iOS developer interview questions will be able to pass the interview on the first try. All the best for your upcoming interviews!

The problem with the given code is that a guard's else block requires an escape path, which can be achieved by RETURNING, throwing an exception, or calling a @noreturn. Adding a return statement is the SIMPLEST approach:

A fatalError() can also be called which is a @noreturn function:

You are given a list of points to cover and the order in which you must do so. Write an ios Swift code to give the shortest NUMBER of steps you can take to ACHIEVE it. You begin from the first point of the given list.

Because the order in which the points are covered is already known, the problem is reduced to figuring out how to calculate the distance between two points (A, B) and the distance between two points (A, B) and the distance between two points (A, B) and the distance between two points (A, B) and the distance between two points (A, B) and the distance (C, D). It is worth noting that just X = ABS(A-C) and Y = abs(A-C) are important (B-D). You will progress along the diagonal while X and Y are both positive, and X and Y will both decrease by one. When one of them reaches zero, you move on to the next stage, reducing the remaining number by one. In other words, the total number of steps would be equal to the maximum number of steps (X, Y)

The code in ios Swift to solve this given problem is given below:

Define and implement a custom power ^^ operator that satisfies the following requirements:

• As arguments, it accepts two Ints.
• The first parameter is returned after raising it to the second parameter's power.
• The EQUATION is correctly evaluated using the conventional algebraic sequence of operations.
• Overflow errors are not taken into account.

There are two steps to creating a new custom operator:

• Declaration: The "operator" keyword is used in the declaration to indicate the type of the operator (unary or BINARY), the sequence of letters that make up the operator, associativity, and precedence. The operator here is ^^ and the type is infix (binary) in this case. Equal precedence operators ^^ should evaluate the equation from right to left (associativity is right). The following is the declaration step:

• Implementation: In Swift, there is no set precedence for exponential operations. Exponents should be calculated before multiplication and division in the normal algebra sequence of operations. As a result, you will need to set custom precedence that puts them ahead of multiplication. The following is the implementation step:

It can be noted that as the program does not take overflows into consideration, in the event of the OPERATION producing a result that Int can't represent, for example, a value more than Int.max, then a runtime error occurs.

Can we do an initialization using the following code:

Yes, Swift SPECIFIES protocols that allow you to use the assignment operator to initialize a type with literal values. Literal initialization of a certain type is possible by adopting the corresponding protocol and supplying a public initializer. You implement ExpressibleByFloatLiteral as follows in the example of Temperature:

After this, we can do the above-given initialization without any errors.

var item = "apples"let closure = { print("He WANTED to EAT \(item)")}item = "oranges"closure()

The above-given code SNIPPET PRINTS "He wanted to eat oranges". If you do not include a capture list in the closure, the compiler will use a reference rather than a copy. As a result, any change to the variable is REFLECTED when the closure is invoked. 

var item = "apples"let CLOSURE = { [item] in print("He wanted to eat \(item)")}item = "oranges"closure()

The output of the given code snippet will be "He wanted to eat apples". When we define the closure, the capture list generates a duplicate of the item. This MEANS that EVEN if you MODIFY the value of an OBJECT, the captured value remains the same.

public class TemperatureClass { private(SET) var temp: Double = 50.0 public func changeTemperature(_ temp: Double) { self.temp = temp }}let temperatureClass = TemperatureClass()temperatureClass.changeTemperature(72.3)public struct TemperatureStruct { private(set) var temp: Double = 50.0 public MUTATING func changeTemperature(_ temp: Double) { self.temp = temp }}let temperatureStruct = TemperatureStruct ()temperatureStruct.changeTemperature(72.3)

Yes, the code given will FAIL to compile because of the last line of the code. The TemperatureStruct has a mutating method to change its INTERNAL variable temp, which is correctly declared. Because we called changeTemperature on an object generated using let, which is immutable, the compiler gives an error. To make the example compile, change let to var.

Methods that affect the internal state of a structure must be MARKED as mutating, but they cannot be invoked from immutable variables.

import UIKitvar viewOne = UIView()viewOne.alpha = 0.7let viewTwo = UIView()viewTwo.alpha = 0.7 // Does this line COMPILE?

The last line will, in fact, compile. viewOne is a variable that we can reassign to a new UIView object. Because we can only assign a value once with let, the FOLLOWING code will not compile:

However, because UIView is a reference-based CLASS, we can change the properties of viewTwo — which means the following code will compile:

struct Course{ var toughness: INT = 3}var courseOne = Course()var courseTwo = courseOnecourseTwo.toughness = 4

What will be the respective values of courseOne.toughness and courseTwo.toughness? If Course was a class instead of a struct, would the values be any DIFFERENT?

The value of courseOne.toughness will be 3 and the value of courseTwo.toughness will be 4 if Course is a structure since Structures in ios Swift are value types and the following line just copies the courseOne to courseTwo by value and not by reference:

If Course was a class instead of a struct, then the above given line would copy courseOne to courseTwo by reference and therefore, the values of both courseOne and courseTwo would be 4 by the end of the code snippet as they have the same ADDRESS. CLASSES in ios Swift are Reference types therefore both the given courses have the same address.