Type | Treat 2021 - Day 2

#typescript #challenge

Type | Treat Challenge 2

Welcome to the second Type | Treat challenge! These challenges are a series of blog posts which have 2 code challenges in, one for beginners and one for intermediate TypeScript programmers. We're on day two, which means going over the answers from yesterday and 2 new challenges.

Yesterday's Solution

Beginner/Learner Challenge

The first part of the solution for this challenge used as const to trigger "Literal Inference" - basically telling TypeScript "Don't convert the array to string[] but consider it a constant set of string literals. This meant that playlist[0] stopped returning string and started returning "The Legend of Sleepy Hollow by The Monotones.mp3".

  const playlist = [
      "The Legend of Sleepy Hollow by The Monotones.mp3",
      ...
- ]
+ ] as const

The second part of the challenge used typeof types to extract the type from the playlist array. Without the first change, this would be string but after the change this meant the full array of different types. You then needed to use the type index syntax [number] to declare that you want any potential string from that array.

- function playSong(song: string) {
+ function playSong(song: typeof playlist[number]) {
      api.play(song)
  }

Successfully completing this challenge would raise an error in the final code samples due to a subtle typo.

Our answer.

Intermediate/Advanced Challenge

This pattern is quite common in code we write in TypeScript codebases, you create one function which takes the result of another and keeps passing objects between functions in a pipeline. One of the best techniques for simplifying this design pattern is to use ReturnType with typeof myFunc to map the return type of one function to the paramter of another. This removes the need for intermediary types which need to be updated when the functions change.

  const findOldCostume = () => {
      // ...
      return { jumpSuit, estimate }
  }

- const createNewMask = (costume: any) => {
+ const createNewMask = (costume: ReturnType<typeof findOldCostume>) => {
      // ...
      return { ...costume, mask }
  }

The little extra step at the end was a small reminder that you can use this technique to provide a type which can be re-used everywhere.

type Costume = ReturnType<typeof assembleCostume>

Our Answer.

The Challenge

Beginner/Learner Challenge

Help sort out a large array of pumpkins, and then get cooking soup.

Intermediate/Advanced Challenge

Fresh back from robotics camp, can you make the perfect punch mixer bot?

How To Share Your Solution

Once you feel you have completed the challenge, you will need to select the Share button in the playground. This will automatically copy a playground URL to your clipboard.

Then either:

Go to Twitter, and create a tweet about the challenge, add the link to your code and mention the @TypeScript Twitter account with the hashtag #TypeOrTreat.
Leave us a comment with your feedback on the dev.to post, or in this post.

Best Resources for Additional Help

If you need additional help you can utilize the following:

The New TypeScript Handbook
The TypeScript Community Discord
The comments on each Dev.to post!
Our previous Type | Treat 2020 challenges

Happy Typing :)

Top comments (7)

cherryblossom000 • Oct 27 '21 • Edited

Lemuel Cushing • Oct 27 '21 • Edited

All the solutions I saw for the begginer challange had either a hardcoded shape for the three pumpkin types, or inferred them from the array, which also locked the Pumpkin type to only the existing combos. (So, no white ripe pumpkin can exist?)

I tried here to create freely combinable version, but it seems to only half worked. First, I had to Omit the soundWhenHit from the base type before I could create the filtering one, not sure why.
Second, the cookPumpkinSoup() still complains. I guess it can't now idetify the initial array as consising of Pumpkins? I'm a bit lost.