Typescript default function parameter based on generic's type

Question

I would like to refactor the following to specify a default implementation for the extractor parameter of the handleEvent method when TEvent is of type DefaultEvent.

Instead of calling:

handleEvent<DefaultEvent>({value: "test"}, (e) => e.value);

I would like the following to be equivalent if possible:

handleEvent<DefaultEvent>({value: "test"});

type DefaultEvent = {
  value: string
}

type SpecificEvent = {
  specificValue: string
}

function handleEvent<TEvent>(
  e: TEvent,
  extractor: (e: TEvent) => string
  // default extractor implementation should be (e) => e.value when TEvent is of type DefaultEvent
  ) {
    console.log(extractor(e));
}

handleEvent<DefaultEvent>({value: "test"}, (e) => e.value);
handleEvent<SpecificEvent>({specificValue: "test2"}, (e) => e.specificValue);

D Pro · Accepted Answer · 2021-04-05 10:29:42Z

1

You can handle the types via function overloading, but you still need to provide the missing extractor is the function implementation. Here's the playground

edited Apr 5, 2021 at 10:29

answered Apr 4, 2021 at 19:35

D Pro

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2 Comments

Dynamic Over a year ago

Thanks - this works well. Is there a way to do this without explicitly having an overload for SpecificEvent? SpecificEvent could be an infinite amount of different events.

Conversion of type 'TEvent' to type 'SpecificEvent' may be a mistake because neither type sufficiently overlaps with the other. If this was intentional, convert the expression to 'unknown' first.ts(2352)

when leaving out extends DefaultEvent | SpecificEvent on the implementation.

D Pro Over a year ago

@Dynamic I have updated the playground link in my answer, please check

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