Elegantly populate dictionary from a struct checking nil values

Question

Given a struct A, I want to populate a NSDictionary with values from that struct, provided they are not nil.
To do that I insert all the valu

Answer 1

It's usually not a good idea to have a dictionary with a value that is optional. Dictionaries use the assignment of nil as an indication that you want to delete a key/value pair from the dictionary. Also, dictionary lookups return an optional value, so if your value is optional you will end up with a double optional that needs to be unwrapped twice.

You can use the fact that assigning nil deletes a dictionary entry to build up a [String : String] dictionary by just assigning the values. The ones that are nil will not go into the dictionary so you won't have to remove them:

struct A {
    var first: String?
    var second: String?
    var third: String?
}

let a = A(first: "one", second: nil, third: "three")

let pairs: [(String, String?)] = [
    ("first", a.first),
    ("second", a.second),
    ("third", a.third)
]

var dictionary = [String : String]()

for (key, value) in pairs {
    dictionary[key] = value
}

print(dictionary)

["third": "three", "first": "one"]

As @Hamish noted in the comments, you can use a DictionaryLiteral (which internally is just an array of tuples) for pairs which allows you to use the cleaner dictionary syntax:

let pairs: DictionaryLiteral = [
    "first":  a.first,
    "second": a.second,
    "third":  a.third
]

All of the other code remains the same.

Note: You can just write DictionaryLiteral and let the compiler infer the types, but I have seen Swift fail to compile or compile very slowly for large dictionary literals. That is why I have shown the use of explicit types here.

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Alternatively, you can skip the Array or DictionaryLiteral of pairs and just assign the values directly:

struct A {
    var first: String?
    var second: String?
    var third: String?
}

let a = A(first: "one", second: nil, third: "three")

var dictionary = [String : String]()

dictionary["first"] = a.first
dictionary["second"] = a.second
dictionary["third"] = a.third

print(dictionary)

["third": "three", "first": "one"]