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When running npm install, it installs @types/chai. Because of this my IDE (webstorm) is using the chai declaration of expect (which does not have toEqual) instead of vitest types. Do we actually need to install those chai types? deleting them seems to fix the issue.

Hey,

Why did you need to cast config.events to any in this line? Can't TS infer the object?

https://github.com/total-typescript/typescript-generics-workshop/blob/main/src/09-reducer.solution.ts#L29

BTW, great repo. Thanks! 😄

The solution here https://github.com/total-typescript/typescript-generics-workshop/blob/main/src/04-generics-advanced/19-generic-interfaces-with-functions.solution.ts gives a Typescript error even though it passes all the tests:

Changing https://github.com/total-typescript/typescript-generics-workshop/blob/main/src/04-generics-advanced/19-generic-interfaces-with-functions.solution.ts#L8 to

clone: <U>(transform: (elem: T | undefined) => U) => Cache<U>;

seems to fix the issue

The provided code in the video does not seem to be working.

https://github.com/total-typescript/typescript-generics-workshop/blob/main/src/02-passing-type-arguments/11-data-fetcher.solution.ts

If you take the code and paste it into Typescript playground (I have done it for you) - you will get a bunch of errors.

The solution is as following:

const fetchData = async <T>(url: string): Promise<T> => await fetch(url).then(resp => resp.json as T)

Hi,

I just spent 1h trying to solve working around partial inference problem and I decided to watch the solution.
I had the right approach to solve it however Matt didn't experience the same error I had in my local IDE.
The 2nd type argument of makeSelectors have a default value {} in the repo which the code in Matt video doesn't.
Once I removed it from my code all the test pass.

thanks!

You have a comment in the 14.5-literal-inference.explainer.ts file:

// When returning the value only, it infers
// the literal type
const returnsValueOnly = <T>(t: T) => {
  return t;
};

const result = returnsValueOnly("a");
//    ^?

This explanation is a bit misleading. In your example, it returns literal type not because of how your function is declared but because of how it is called. In other words, it is returned literal because "a" parameter is interpreted as literal.

For example, if you pre-declare the function's argument as let the type of return value will be a string: