Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

iOS Swift 2.0 Enumerations and Optionals Introduction to Optionals Initializing Optional Values

Posted by jc laurent
jc laurent
6,351 Points

Make sure you are creating a failable initializer that accepts a dictionary as an argument for initialization

I thought I did it??

optionals.swift 
struct Book {
    let title: String
    let author: String
    let price: String?
    let pubDate: String?

    init?(withDictionary dict: Dictionary<String, String>) { // See comment below!
        guard let title = dict["title"] else { return nil }
        self.title = title

        guard let author = dict["author"] else { return nil }
        self.author = author

        self.price = dict["price"]
        self.pubDate = dict["pubDate"]
    }
}

5 Answers

Moritz Lang
Moritz Lang
25,909 Points
Moritz Lang
Moritz Lang
25,909 Points

This is my solution:

init?(dict: [String: String]) {
        // non-optional values
        guard let title = dict["title"] else { return nil }
        guard let author = dict["author"] else { return nil }
        self.title = title
        self.author = author
        // optional values
        self.price = dict["price"]
        self.pubDate = dict["pubDate"]
    }
Moritz Lang
Moritz Lang
25,909 Points
Moritz Lang
Moritz Lang
25,909 Points

Hi, why do you check the constants that arent optional and not the ones that are optional? :)

By the way: Dictionary<String, String> is not the best way to declare a dictionary of type String: String. It could be easily written like [String: String].

Moritz Lang
Moritz Lang
25,909 Points
Moritz Lang
Moritz Lang
25,909 Points

Oh, I'm sorry. I interpreted your code wrong. Just give me a moment and I'll provide you with a better solution :)

jc laurent
jc laurent
6,351 Points
jc laurent
jc laurent
6,351 Points

thks, but that's still giving this error = "Make sure you are creating a failable initializer that accepts a dictionary as an argument for initialization"

Moritz Lang
Moritz Lang
25,909 Points
Moritz Lang
Moritz Lang
25,909 Points

That's very strange. I can pass this challenge with exactly the code I've written above. Maybe you need to reload the challenge? :)

jc laurent
jc laurent
6,351 Points
jc laurent
jc laurent
6,351 Points

ok Bud

jc laurent
jc laurent
6,351 Points
jc laurent
jc laurent
6,351 Points

you were right! Don't know why it freezes