Can I use named TS types in generics same as named parameters in JS?

Question

TS introduced default parameters types for generics in v2.3 Generic parameter defaults

So I can define a generic type as follows:

interface MyGenericType<T = any, U = any, V = any, W = any> { }

And then I can use it as follows:

declare const xyz: MyGenericType<any, any, any, Date>;

I am trying to get rid of these default types and only specify the ones I want to define by names as follows?

declare const xyz: MyGenericType<{ W: Date }>;

However that did not work as expected because { W: Date } is replacing T the first param not replacing W.

Is that thing not doable in Typescript?

Is there any workarounds other than defining any in the generic type declaration?

Praveenkumar · Accepted Answer · 2020-09-05 06:37:55Z

2

This is a work around. You can create another interface and use it like this.

interface MyGenericType<T = any, U = any, V = any, W = any> { }

interface MyExtendedGenericType<W> extends MyGenericType<any, any, any, W> { }

declare const xyz: MyExtendedGenericType<Date>;

answered Sep 5, 2020 at 6:37

Praveenkumar

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Ingo Bürk · Accepted Answer · 2020-09-05 06:46:41Z

2

Partially inferring generic type arguments is not currently possible. The discussion for this feature can be found here:

https://github.com/microsoft/TypeScript/issues/26242

In there you'll also find the link to a now closed issue that discussed referring to the type arguments by name (as in your example) which was closed because this is a rather big change as it would make the name of the type argument external to the type itself, which has rather big implications.

answered Sep 5, 2020 at 6:46

Ingo Bürk

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