Swift Optional Dictionary [String: String?] unwrapping error

Question

So here I have a basic setup

var preferenceSpecification = [String : String?]()
preferenceSpecification["Key"] = "Some Key"
preferenceSpecification["Some Key"] = nil
preferenceSpecification["DefaultValue"] = "Some DefaultValue"
print(preferenceSpecification)
var defaultsToRegister = [String : String]()

if let key = preferenceSpecification["Key"], let defaultValueKey = preferenceSpecification["DefaultValue"] {
    defaultsToRegister[key] = preferenceSpecification[defaultValueKey]!
}

But the error points out where it demands that I force unwrap this, to be like this:

defaultsToRegister[key!] = preferenceSpecification[defaultValueKey!]!

Which doesn't make sense, because keyValue and defaultValue already are unwrapped

Possible duplicate of [Optional Type Dictionary yields "Value of Optional type not unwrapped"](http://stackoverflow.com/questions/29691994/optional-type-dictionary-yields-value-of-optional-type-not-unwrapped) — Hamish, Aug 18 '16 at 19:43
Compare http://stackoverflow.com/questions/27225232/two-or-more-optionals-in-swift: `preferenceSpecification["..."]` is a `String?` because that is the value type of `[String : String?]()`. Do you really want the value to be optional? What should happen if `preferenceSpecification["Key"]` is defined as `nil` in the dictionary? — Martin R, Aug 18 '16 at 19:55
It's expected a different action, that node is ignored. but as @Hamish pointed out, and the question you marked closes the case. There's no need to mark [String: String?], the value is optional by default. — RodolfoAntonici, Aug 18 '16 at 19:59

Luca Angeletti · Accepted Answer · 2016-08-18T20:16:26.580

When you extract a value from a dictionary like this using subscript

[String: String?]

you need to manage 2 levels of optional. The first one because the subscript returns an optional. The second one because the value of you dictionary is an optional String.

So when you write

if let value = preferenceSpecification["someKey"] {

}

you get value defined as an optional String.

Here's the code to fix that

if let
    optionalKey = preferenceSpecification["Key"],
    key = optionalKey,
    optionalDefaultValueKey = preferenceSpecification["DefaultValue"],
    defaultValueKey = optionalDefaultValueKey,
    value = preferenceSpecification[defaultValueKey] {
    defaultsToRegister[key] = value
}

Suggestions

You should avoid force unwrapping as much as possible. Instead you managed to put 3 ! on a single line!
You should also try to use better name for your constants and variables.

It was just a test case to place it on the StackOverflow, the code on the project is way more complex. But as it was pointed out, the value of an object on dictionary may be nil, it's by default an optional. A little deeper: https://developer.apple.com/swift/blog/?id=12 — RodolfoAntonici, Aug 18 '16 at 20:02

score 1 · Answer 2 · answered Aug 05 '20 at 11:52

You could also define an extension which helps get rid of the double optional situation.

extension Dictionary where Value == Optional<String> {
    
    func flattened(_ key: Key) -> Value {
        if let value = self[key] {
            return value
        }
        return nil
    }
        
}

Usage: preferenceSpecification.flattened("someKey")

Swift Optional Dictionary [String: String?] unwrapping error

2 Answers2

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