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  
}