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Slice Sizing

Slice Sizing

Learning Exercise

Introduction

Optionals

Swift uses optionals to allow programmers to represent the possible absence of a value. Optional is a type that can either hold a value or be nil, which represents the absence of a value. Using an optional requires a program to check if a value does exist before using it after unwrapping it.

Any type can be made into an optional by appending a ? onto the end of the type name. So an optional integer would have type Int? and an optional string would have type String?. Defining constants or variables of optional type and assigning them values is done the same as for values of non-optional types.

let x: Int? = 42
var y: String? = "Hello"
y = "Goodbye"

You can assign the absence of a value to a variable of optional type by assigning it the special value nil. nil can be used with all optional types, but nils assigned to two different optional types do not have the same type, and cannot be interchanged or even compared.

let intOpt: Int? = nil
let stringOpt: String? = nil

intOpt = stringOpt
// Compiler error: Cannot assign value of type 'String?' to type 'Int?'

intOpt == stringOpt
// Compiler error: Binary operator '==' cannot be applied to operands of type 'Int?' and 'String?'

Also note that even though nil can be assigned to any optional type, it cannot be assigned to a non-optional type (even if it doesn't actually hold nil). And methods that is expecting a non-optional type cannot be passed an optional type without unwrapping it first.

var x: Int = 42
var y: Int? = 42

y = x
// Works fine
x = y
// error: value of optional type 'Int?' must be unwrapped to a value of type 'Int'
x + y
// error: value of optional type 'Int?' must be unwrapped to a value of type 'Int'

Using optionals

Because optional types are not the same types as their base types, the two types cannot be used in the same ways. For example: Int("123") + 1 results in a compiler error "Value of optional type 'Int?' must be unwrapped to a value of type 'Int'". This is because the Int("123") returns an optional Int? type, not an Int type, since if the string cannot be converted to an integer, the result will be nil. In order to access the Int from the conversion, one must "unwrap" it first.

This is most commonly done in Swift using the if-let and guard-let constructs for optional binding which check for nil and take one code path with the unwrapped value bound to a supplied name if a value exists and taking a different code path if nil was found.

if let safeNum = Int("123") {
	let sum = safeNum + 1
	…
} else {
  // code for the case where nil was found -- may be left out
}

It is worth noting that the safeNum variable has the type Int and not Int?. In the example below, num is of type Int? and safeNum is of type Int.

let num = Int("123")
if let safeNum = num {
  // num is of type Int
}

This optional binding is important because it unwraps (or "removes") the optional type from the value, allowing it to be used as a non-optional value. If you would just do a conditional check on if the value is nil, the value would still be of optional type:

let num = Int("123")
if num != nil {
  // num is of type Int?
}

The guard-let option may also be used in the cases where early return is desired:

guard let num = Int("123") else { return nil }
let sum = num + 1
…

Comparing optionals

Note that even if the base type of a pair of optionals can be compared using the standard comparison operators, the optionals themselves cannot be compared. They can only be checked for equality. Two optionals are equal if they are both nil or if the values they wrap are equal within their base types.

However, code can be written to perform a custom comparison of two optional values. Below is an example of a switch statement that will return true only if both optional values are non-nil and the first value is less than the second. To do this it uses the optional pattern varName? which only matches non-nil optionals, binding the value inside the optional to the name varName.

switch (optionalA, optionalB) {
case let (valA?, valB?): return valA < valB
default: return false
}

Instructions

You have a number of pizza slice shops in your town and you want to write a webapp that will let you compare two different pizza configurations to let you know who will give you the bigger slice.

1. Write a function to compute slice sizes which returns nil for invalid input.

Implement the function, sliceSize(diameter: Double?, slices: Int?) -> Double?, which, given the diameter of a pizza and the number of slices per pizza returns the area of a slice. For negative diameters and for number of slices less than 1, return nil, as there is no such thing as a pizza with negative diameter and no way to slice a pizza into fewer than 1 slice. If either parameter is nil, also return nil

sliceSize(diameter: 16, slices: 12)
// Returns 16.75516081914556
sliceSize(diameter: nil, slices: 8)
// Returns nil

2. Process input from your web application to determine the larger slice.

You web application will pass four strings to your function, biggestSlice(diameterA: String, slicesA: String, diameterB: String, slicesB: String) -> String. The first and second strings are the diameter and number of slices of the first pizza respectively, and the third and fourth are the diameter and number of slices of the second pizza respectively.

Implement biggestSlice so that it attempts to convert the diameter and number of slices for each pizza into a Double and an Int respectively. If both of these values can be obtained from the strings, use your first function to try to compute the area, otherwise the area for that slice is nil. Once the areas of both slices are obtained, compare the two areas using the following rules:

  1. If slice A's area is a Double and slice B's is nil, return "Slice A is bigger". If the reverse is true, return "Slice B is bigger".
  2. If both areas are Doubles, return "Slice A is bigger" or "Slice B is bigger" according to which slice has the greater area.
  3. If both areas are nil, or if both are Doubles and they are equal, return "Neither slice is bigger".
Note

Even if the input result in a slice area of 0, the slice is still considered valid, and should be seen as bigger than a slice with an area of nil.

biggestSlice(diameterA: "10", slicesA: "6", diameterB: "14", slicesB: "12")
// Returns "Slice A is bigger"
biggestSlice(diameterA: "10", slicesA: "6", diameterB: "12", slicesB: "8")
// Returns "Slice B is bigger"
biggestSlice(diameterA: "Pepperoni", slicesA: "6", diameterB: "Sausage", slicesB: "12")
// Returns "Neither slice is bigger"
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