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iOS Swift Functions and Optionals Parameters and Tuples Decomposing a Tuple

Posted by Piyawit Wipoosiri
Piyawit Wipoosiri
Courses Plus Student 7,122 Points

In the extra credit of "Parameters and Tuples", What is a syntax for a function that has Dictionary as parameter.

I can't figure out how to write the function that you give in the extra credit. Would you please show the correct answer or example.

2 Answers

Dan Chambers
Dan Chambers
6,057 Points
Dan Chambers
Dan Chambers
6,057 Points

This is what I came up with. It took me ages I must admit. I found it to be more of a syntax/documentation challenge than anything.

import UIKit

func nowPlaying (#song: Dictionary<String,String>) -> (title: String, artist: String, album: String) {
    let title = song["Title"]
    let artist = song["Artist"]
    let album = song["Album"]

    return (title: title!, artist: artist!, album: album!)
}


let song1 = [   "Title":"Singing in The Rain",
                "Artist":"Frank Sinatra",
                "Album":"Best of Frank"]


let currentSong = nowPlaying(song: song1)

println("The song that is currently playing is \(currentSong.title) by \(currentSong.artist) from the album \(currentSong.album).")
Kishan Solomon
Kishan Solomon
1,135 Points
Kishan Solomon
Kishan Solomon
1,135 Points

What do the bangs(exclamation marks) do for the nowPlaying returns? i.e. title: title! instead of just title:title.

Robin Nilsson
Robin Nilsson
3,783 Points
Robin Nilsson
Robin Nilsson
3,783 Points

The exclamation marks seem to be required (just using "title: title," produces an error in Xcode). The syntax is introduced for Optionals in the next video in the course. Does this mean that the individual values in a Tuple is actually counted as Optionals and need to be unwrapped just like Optionals do?

Austin Langley
Austin Langley
5,392 Points

Answer to:

What do the bangs(exclamation marks) do for the nowPlaying returns? i.e. title: title! instead of just title:title.

I had the same question as Kishan Solomon. I found the answer online:

let title = song["Title"]
let artist = song["Artist"]
let album = song["Album"]

The constants that are being created here will be automatically typed as optional by Swift because the keys may or may not exist in the dictionary. We use the bang to unwrap them.

I hope that helps.

Stone Preston
Stone Preston
42,016 Points
Stone Preston
Stone Preston
42,016 Points

there are two styles of stating dictionary types.

the swift eBook section on collection types states:

The type of a Swift dictionary is written in full as 

Dictionary<Key, Value>

, where Key is the type of value that can be used as a dictionary key, and Value is the type of value that the dictionary stores for those keys.

so we could create a function that takes a dictionary parameter with strings as its values and keys like so:

func someFunct(dict: Dictionary<String: String>) {

}

the eBook then states:

You can also write the type of a dictionary in shorthand form as [Key: Value]. Although the two forms are functionally identical, the shorthand form is preferred

so could create the same function using shorthand syntax for the parameter

func someFunct(dict: [String: String]) {

}
Piyawit Wipoosiri
Piyawit Wipoosiri
Courses Plus Student 7,122 Points

Thank you for your answer.... I try to create a function that way and there is no error. But I still try to call that function by passing a Dictionary Parameter in it, but always error. (I tried in Playground, Xcode) I also still find a sample syntax to use Dictionary Parameter which put in the function, e.g. println() but nothing happen...?

Thank you very much.