// Swift Quiz

struct Person {
let name: String

mutating func changeName() {
// ...
}
}

var person = Person(name: "Florian")
person.changeName()

// Can you implement `changeName` so that it changes `https://t.co/Lp6nga6UC6`?

Answer: Yes, in plain Swift

While you can't mutate `name` directly, you *can* reassign `self`:

mutating func changeName() {
self = Person(name: "Chris")
}

// Swift Quiz
// by @tammofreese

import UIKit

let color1 = UIColor(white: 1 / 255, alpha: 1.0)
let color2 = UIColor(white: 1.0 / 255.0, alpha: 1.0)

let result = (color1 == color2)

// What‘s the value of result?

Solution: true

Because this initializer of UIColor takes a CGFloat, the expression 1 / 255 gets typed as a CGFloat expression, and will have the same value as 1.0 / 255.https://t.co/s7500CWmAg

// Swift Quiz
// by @tammofreese

func foo() throws -> Int { 0 }
func foo() throws -> Int? { 1 }

let a = (try? foo())
let b = (try? foo())!

let result = (a == b)

// What’s the value of result? (Swift 5)

Solution: true

In most other cases (e.g. when you write `try foo()`), Swift complains about an ambiguous use of `foo()`. However, with a `try?` it chooses the optional overload.https://t.co/s7500CWmAg

// Swift Quiz

var language = "Objc"

let code = { [language] in
print(language)
}

language = "Swift"

let newCode = code
newCode()

// What's the output?

Solution: Objc

The moment the closure is created (when we define `code`) the `language` variable is captured: the value at that point in time is available within the closure's body.https://t.co/s7500CWmAg

// Swift Quiz

class X {
lazy var a: Int? = {
print("A")
return nil
}()
func b() -> Int? {
print("B")
return a ?? 3
}
let c: () -> Int = {
print("C")
return 5
}
}
let x = X()
x.b() ?? x.c()

// What prints?

Solution: BA

When `x.b()` is called, "B" gets printed.

Within `b`, `a` is accessed, which executes the code in the closure of the lazy stored property, printing "A".

Since `b` returns a non-`nil` value, `c` is never called.https://t.co/s7500CWmAg