How to unwrap an optional dictionary value in Swift 4+ in one step?

Question

Given the dictionary of dictionary below, what is the correct syntax to unwrap the Int? in one step?

let dict:Dictionary<String, Dictionary<String, Int?>> = [
"parentKey" : [
    "firstKey" : 1,
    "secondKey" : nil]
]

let x = "someKey"
let y = "someOtherKey"
var foo = 0

if let goo = dict[x]?[y] { foo = goo } //<-- Error: cannot assign (Int?) to Int

if let goo = dict[x]?[y], let boo = goo { foo = boo } //<-- OK

In the first 'if let', goo is returned as an Int? - goo then needs to be unwrapped as in the second 'if let'...

What is the correct syntax for doing this in one step?

This gives the correct result when both keys are valid (e.g. x = "parentKey, y = "firstKey" or y = "secondKey". However, it gives an incorrect result when either x or y is a non-existent key (e.g. x = "someKey" or y = "someOtherKey") — Andrew Coad, Dec 01 '18 at 05:11

score 1 · Answer 1 · answered Nov 23 '18 at 04:38

1

Use nil colesecing and provide a default value. Only way to safely unwrap a dictionary value.

if let goo = dict[x]?[y] ?? NSNotFound { foo = goo }

answered Nov 23 '18 at 04:38

SirCJ

505
6
11

score 1 · Accepted Answer · answered Nov 23 '18 at 05:20

As far as I understood you want to force unwrap a double optional. There different methods.

let dbOpt = dict[x]?[y]

My favorite:

if let goo = dbOpt ?? nil { foo = goo }

Using flatMap:

if let goo = dbOpt.flatMap{$0} { foo = goo }

Using pattern matching:

if case let goo?? = dbOpt { foo = goo }

score 0 · Answer 3 · answered Nov 23 '18 at 10:25

0

There are many ways to do this but one of the simplest solution available is :

var foo = 0

if let goo = dict[x]?[y]  as? Int{
    foo = goo
}
print(foo)

answered Nov 23 '18 at 10:25

nikksindia

1,147
1
10
18

How to unwrap an optional dictionary value in Swift 4+ in one step?

3 Answers3