爲什麼隱式解包可選未分配？

Question

在下面的代碼中，爲什麼5沒有賦值給「somevar」？

class ViewController: UIViewController { 

    var somevar : Int? 

    override func viewDidLoad() { 
     super.viewDidLoad() 
     somevar! = Int(5) // why 5 is not assigned to somevar here 
    } 
} 

背景：

somevar被聲明爲可選變量，這意味着如果該變量是零則命令使用此變量將被忽略。

例子：

class ViewController: UIViewController { 

    var somevar : Int? 

    override func viewDidLoad() { 
     super.viewDidLoad() 
     somevar? = 5 // this command will be ignored bcz somevar is nil 
    } 
} 

要在我們自己的風險強行執行的命令，我們使用「隱式展開可選」，使我們確信，命令將被執行被執行，在這種情況下，下面一行

somevar! = 5 

fatal error: unexpectedly found nil while unwrapping an Optional value

在執行這行，爲什麼「5」不分配給「somevar」，而不是出現致命錯誤？

class ViewController: UIViewController { 

    var somevar : Int? 

    override func viewDidLoad() { 
     super.viewDidLoad() 
     somevar! = 5 
    } 
} 

Answer 1

當我們做something!（強調！馬克），我們是 「力閱讀」（力展開）可選。

也就是說，在爲其分配新值之前，上面的代碼嘗試讀取something。
由於something是nil，代碼爆炸。

爲了說明：

var somevar: Int? 

print(somevar!) 
// Code explodes! 

print(somevar) 
// Output is "nil" 

somevar = 5 

print(somevar!) 
// Output is "5" 

print(somevar) 
// Output is "Optional(5)" 

作爲@LeoDabus說明，本是覆蓋在Apple's awesome Swift book。
（BTW一個很好的書！❤️）

Answer 2

給予一定的顏色是什麼somevar? = 5在做什麼。

//: Playground - noun: a place where people can play 

import Foundation 

// In swift you have to unwrap an optional before you can do anything with it 
var x: Int? = 1 
var y: Int? = 2 

// So you can't do this 
//var z = x + y 

// You have to do this 
if let x = x, 
    let y = y { 
    // Here x and y are no longer of the type Int? they are of the type Int 
    var z = x + y 
} 

// You don't have to name them the same 
if let someX = x, 
    let someY = y { 
    // Here x and y are no longer of the type Int? they are of the type Int 
    var z = someX + someY 
} 

// This can be a pain sometimes if you want to "do nothing" in the nil case, or want to unwrap something multiple "levels" 
// of optionals deep. For example: 

struct Pet { 
    let name: String 
} 

struct Person { 
    let pet: Pet? 
} 

var person: Person? = Person(pet: Pet(name: "Rex")) 

// To get the person's pet's name we have to unwrap a few things 
if let person = person, 
    let pet = person.pet { 
    print("The pet's name is \(pet.name)") 
} 

// We can do this a little easier by using "Optional Chaining" 
if let name = person?.pet?.name { 
    print("The pet's name is \(name)") 
} 

// So here's where your problem comes in 
var number: Int? = nil 

// This is "optional chaining" the assignment of 5 to number. But, because number is current nil the assignment won't happen. 
number? = 5 

// However 
number = 5 

// Now the number is 5 

number? = 10 

// Now the number is 10, because the optional chaining succeded because number was not nil. All this being said, I've never 
// seen someone use x? = 5 in production code, and I can't think of a reason to do that. Just do x = 5 like the other 
// answers have said. 

TL; DR，somevar? = 5使用可選的鏈接，只設置somevar到5如果不是nil。