"For-in loop requires '[String]?' to conform to 'Sequence'; did you mean to unwrap optional?" but I don't think I am using optionals

Question

Here is what I'm trying to do but simplified down:

var adictionary = [String:[String]]()
adictionary["A"] = ["B", "C"]
adictionary["B"] = ["A", "C"]
adictionary["C"] = ["A"]
adictionary["D"] = []

var newdic = Array(adictionary.keys).reduce(into: [String: Bool]()) { $0[$1] = false } //I make an arr from a dictionary's keys of type [String:[String]]

for (key, val) in newdic{
     for arr in adictionary[key]{
          if !(adictionary[arr].contains(key)){
               newdic[key] = true
          }
     }
}

I get these errors, when I run the above simplified version of the code I am trying to run in an ios app:

main.swift:12:28: error: value of optional type '[String]?' must be unwrapped to a value of type '[String]' for arr in adictionary[key]{

main.swift:12:28: note: coalesce using '??' to provide a default when the optional value contains 'nil' for arr in adictionary[key]{

main.swift:12:28: note: force-unwrap using '!' to abort execution if the optional value contains 'nil' for arr in adictionary[key]{

I don't understand, what is wrong with what I am doing? What do these errors mean? It seems like Swift thinks I have an optional somewhere? But I don't see how I do... Any help is much appreciated.

If I change this line: for arr in adictionary[key]{

To: for arr in adictionary[key]!{

It fixes the issues and new errors appear on the if !(a... line. I still don't understand why that fixed it.

Answer 1

Swift dictionary accesses always return optional values, because the lookup will return nil if the key doesn't exist. It is up to you to handle this resulting optional is a safe way. Adding ! is rarely the right way to fix it because your code will crash if nil is returned.

The first time you get an optional is here:

for arr in adictionary[key] {

Again, Swift doesn't know if key exists in adictionary so it returns an optional. A safe way to fix this is to use the dictionary lookup which returns a default value when the key doesn't exist. In this case, returning an empty array seems like a good choice since your for loop will then just do nothing:

for arr in adictionary[key, default:[]] {

Next, you get an optional here:

if !(adictionary[arr].contains(key)){

Again, you need to decide how to safely handle the fact that dictionary[arr] could return nil. The same trick works here: return an empty array if arr doesn't exist:

if !(adictionary[arr, default:[]].contains(key)){

---

Here is the final version of your code:

var adictionary = [String:[String]]()
adictionary["A"] = ["B", "C"]
adictionary["B"] = ["A", "C"]
adictionary["C"] = ["A"]
adictionary["D"] = []

var newdic = Array(adictionary.keys).reduce(into: [String: Bool]()) { $0[$1] = false } //I make an arr from a dictionary's keys of type [String:[String]]
print(newdic)

for key in newdic.keys {
    for arr in adictionary[key, default:[]] {
        if !(adictionary[arr, default:[]].contains(key)){
               newdic[key] = true
          }
     }
}