How does one type this default value in TypeScript?

Question

Sorry for the vague title, but I don't know how to express this problem concisely.

Basically, I have the following code (with unimportant bits removed):

function identity<T>(value: T): T {
    return value;
}

function extent<T, K>(
    values: T[],
    map: (value: T) => K = identity // the '= identity' part gives an error
): [K, K] | undefined {
    let min;
    let max;

    // ... stuff ...

    return min === undefined || max === undefined ? undefined : [map(min), map(max)];
}

Where the default = identity value results in a compiler error.

I can remove the default value for the map parameter from the extent's signature and always provide one myself:

extent([5, 1, 3, 4, 8], x => x)

Which will work just fine, but I'd rather not supply the second parameter, unless I'm actually mapping the values from one type to another.

How do I type the indentity function so that it's accepted as the default value for extent's map parameter? I'm stuck on TypeScript 3.6 for the time being, if that matteres.

spender · Accepted Answer · 2021-09-08 11:33:08Z

1

Probably the best way to deal with this that maintains type-safety to the outside world is via overloads:

function extent<T>(values: T[]): [T, T] | undefined
function extent<T, K>(values: T[], map: (value: T) => K): [K, K] | undefined
function extent<T, K>(
  values: T[],
  map?: (value: T) => K 
): [K, K] | [T, T] | undefined {
  const actualMap = map ?? identity;
  let min;
  let max;

  // ... stuff ...

  return min === undefined || 
         max === undefined ? 
             undefined : 
             [actualMap(min), actualMap(max)];
}

edited Sep 8, 2021 at 11:33

answered Sep 7, 2021 at 17:45

spender

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3 Comments

Vitaly Over a year ago

Awesome! That’s the no compromise solution I’ve been looking for! Thanks 😊

Vitaly Over a year ago

Actually, it still fails if I declare the type of min/max: let min: T; let max: T; I mean, in my code I don't have to declare it because it is inferred by the compiler from the code section I omitted (marked as "... stuff ..."), but it fails nontheless. So the return type of extent should be not [K, K] | [T, T] | undefined but [T | K, T | K] | undefined, but this solution is still valid 👍🏻

spender Over a year ago

@Vitaly I had the same identity function with the same signature in some of my own projects, and it took a short while to get this clean (or at least, clean for any consumers!). Glad to help.

Phat M Tran · Accepted Answer · 2021-09-07 17:03:23Z

0

It's because of type error.

In map parameter you declare that this function will return the type of K but in the identity function you expect that to return the type of T so you can't assign it as a default value.

You could fix it like this: map: (value: K) => K = identity

answered Sep 7, 2021 at 17:03

Phat M Tran

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2 Comments

spender Over a year ago

That's not going to be much use if the mapping is to something other than K.

Phat M Tran Over a year ago

Yeah, I just quickly answer in the author's case.

udalmik · Accepted Answer · 2021-09-07 17:22:56Z

0

There is a conflict with T => K type of map and K => K type in identity function. Making identity less strict on types can help:

function identity<T>(value: T): any {
    return value;
}

or even

const identity = (x: any) => x;

answered Sep 7, 2021 at 17:22

udalmik

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1 Comment

Vitaly Over a year ago

Yes, but then the compiler won’t be able to properly infer the return type of the extent function. It would be either [unknown, unknown] or [any, any].

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