CODEWITHSUNDEEP
swift
snaggs
snaggs

Reputation: 5713

Why my optional value is not unwrapped, Swift?

I have pretty simple example and have no clue why it doesn't work as expected.

var list:Array<Int> = [1,2,3,4,5]

var item:Int?


for var index = 0; index < list.count; index++ {

   item! = list[index]

   item = item + 5 // <-- error value of optional type 'Int?' not unwrapped
}

Why Swift forces me to write: item = item! + 5

I unwrapped it here: item! = list[index] and if list returns nil - the Exception will be thrown.

As I understand on this step a.e.: item! = list[index] the item is not nil

I tried several options like:

But still get the same demand to write item = item! + 5

I use playground

Upvotes: 0

Views: 620

Answers (1)

Vatsal
Vatsal

Reputation: 18171

Let me break it down for you:

  • var item:Int? tells the compiler that whenever the word item is used, it refers to a variable of type optional Int (aka Int?).

  • var item:Int on the other hand, tells the compiler that whenever the word item is used, it refers to a variable simply of type Int.

  • In order to access the unwrapped value of your variable declared in var item:Int?, you will always have to use item!. The compiler is not going to guess whether the variable item has a value or not. That, after all, is the the whole purpose of optionals. To make it clear that these kind of variables may or may not have a value.

  • Essentially, what I'm trying to say is that a variable once declared as an optional, will always be an optional regardless of whether it contains a value or not. To get the unwrapped value, you will always have to use the character !, unless you decide to store it's value in another variable (ex. var unwrappedItem = item!)

To get rid of your error, simply declare your item variable to be of type Int, and not Int?.

As to why Swift throws an error instead of just letting the runtime raise an exception, it's just an extra precaution to probably discourage people from using bad practice.

Upvotes: 3

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