TypeScript generics and function parameter conditional types

Question

Let's say I have a function that returns the return value of a passed-in getIdentifier function. This function is mandatory, except for when the item has an id property because I can fall back to reading just that.

type FuncArg<T> = T extends { id: string }
  ? { item: T; getIdentifier?: (item: T) => string }
  : { item: T; getIdentifier: (item: T) => string };

type withId<T> = T & { id: string };

const getFoobar = <T>(arg: FuncArg<T>) => {
  return arg.getIdentifier ? arg.getIdentifier(arg.item) : (arg.item as withId<T>).id;
};

All well so far. Let's say I have a second function that passes in an item to the first function, checks if it has an id and if not it passes in a random generator getIdentifier.

const getFoobar2 = <T>(item: T) => {
  if ('id' in item) return getFoobar({item: item as withId<T>});
  return getFoobar({item: item, getIdentifier: () => Math.random().toString()});
}

Unfortunately the typechecking breaks in the last example and I have no idea to fix this. I've tried everything from conditional types to union types and function overloads. Everytime I run into similar problems. What am I missing here?

Playground ▶

Yehor Androsov · Accepted Answer · 2020-07-21 11:34:40Z

1

You missed item parameter in getIdentifier function signature and looks like it requires explicit as FuncArg<T> in object declaration. Can't explain better, don't have much experience in Typescript.

const getFoobar2 = <T>(item: T) => {
  if ('id' in item)
    return getFoobar({ item: item } as FuncArg<T>);
  return getFoobar({item: item, getIdentifier: (item: T) => Math.random().toString()} as FuncArg<T>);
}

answered Jul 21, 2020 at 11:34

Yehor Androsov

6,1862 gold badges30 silver badges56 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Gersom Over a year ago

That's not it unfortunately. I tried that. If you click on 'Playground ▶' you can play around with the error.

Yehor Androsov Over a year ago

@Gersom typescriptlang.org/play/#code/…

Gersom Over a year ago

Thanks for having a look with me! Technically this works, but I still feel like there is a gotcha that I'm missing, since it's quite verbose / feels unnecessary.

jcalz Over a year ago

This is the right procedure although the answer should maybe be edited to explain what's happening: the compiler cannot usually verify assignability to unspecified generic types, especially conditional types that depend on unspecified generic type parameters. If T were some specific type, the compiler could evaluate FuncArg<T>, but since it's a generic type parameter, the compiler defers evaluating it until later. So it won't know if {item: item}is or is not assignable because it doesn't know what T is. There are github issues about this, let me see if I can find some

jcalz Over a year ago

There's microsoft/TypeScript#36737 and microsoft/TypeScript#35257 and microsoft/TypeScript#32990...

Collectives™ on Stack Overflow

TypeScript generics and function parameter conditional types

1 Answer 1

5 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

5 Comments

Your Answer

Sign up or log in

Post as a guest

Related