Why print() is printing my String as an optional?

Question

I have a dictionary and I want to use some of its values as a key for another dictionary:

let key: String = String(dictionary[\"anotherKey\"])

Answer 1

A Swift dictionary always return an Optional.

dictionary["anotherKey"] gives Optional(42), so String(dictionary["anotherKey"]) gives "Optional(42)" exactly as expected (because the Optional type conforms to StringLiteralConvertible, so you get a String representation of the Optional).

You have to unwrap, with if let for example.

if let key = dictionary["anotherKey"] {
    // use `key` here  
}

This is when the compiler already knows the type of the dictionary value.

If not, for example if the type is AnyObject, you can use as? String:

if let key = dictionary["anotherKey"] as? String {
    // use `key` here  
}

or as an Int if the AnyObject is actually an Int:

if let key = dictionary["anotherKey"] as? Int {
    // use `key` here  
}

or use Int() to convert the string number into an integer:

if let stringKey = dictionary["anotherKey"], intKey = Int(stringKey) {
    // use `intKey` here  
}

Answer 2

This is by design - it is how Swift's Dictionary is implemented:

Swift’s Dictionary type implements its key-value subscripting as a subscript that takes and returns an optional type. [...] The Dictionary type uses an optional subscript type to model the fact that not every key will have a value, and to give a way to delete a value for a key by assigning a nil value for that key. (link to documentation)

You can unwrap the result in an if let construct to get rid of optional, like this:

if let val = dictionary["anotherKey"] {
    ... // Here, val is not optional
}

If you are certain that the value is there, for example, because you put it into the dictionary a few steps before, you could force unwrapping with the ! operator as well:

let key: String = String(dictionary["anotherKey"]!)

Answer 3

You are misunderstanding the result. The String initializer does not return an optional. It returns the string representation of an optional. It is an non-optional String with value "Optional(42)".

Answer 4

You can also avoid force unwrapping by using default for the case that there is no such key in dictionary

var dictionary = ["anotherkey" : 42]

let key: String = 
String(dictionary["anotherkey", default: 0])
print(key)