CODEWITHSUNDEEP
javascripttypescriptramda.js
knpwrs
knpwrs

Reputation: 16496

How can I generically type variable bindings in TypeScript?

I have a generic interface available to me as such:

CurriedFunction3<T1, T2, T3, R>: R

I would like to create a const binding that fills in some of those type arguments. Something like this:

type CurriedSorter<T> = CurriedFunction3<(obj: T) => any, string, T[], T[]>;

When I try to assign a const binding this type, however, I get an error:

type CurriedSorter<T> = CurriedFunction3<(obj: T) => any, string, T[], T[]>;

const sortFactory: CurriedSorter = uncurryN(3, flip(compose(
  unapply(sortWith),
  uncurryN(2, ifElse(
    compose(equals('DESC'), toUpper),
    always(descend),
    always(ascend),
  )),
)));

Generic type CurriedSorter requires 1 type argument(s).

The const binding sortFactory should be a function with one generic type parameter, to be used as such:

sortFactory<MyType>(
  prop('name'),
  'DESC',
  [{ name: 'foo' }, { name: 'bar' }]
) // returns `[{ name: 'bar' }, { name: 'foo' }]`

How can I generically type variable bindings in TypeScript? Is there any way to do this with TypeScript?

Upvotes: 5

Views: 381

Answers (1)

Ryan Cavanaugh
Ryan Cavanaugh

Reputation: 221212

You can't have a bare T in a variable declaration's type because the T could potentially leak out of the value through a property. However, you can have a generic call signature as part of the type of the variable. Here's a self-contained example since I don't know what you've defined uncurryN, unapply, etc as:

type CurriedFunction3<T1, T2, T3, R> = (a: T1, b: T2, c: T3) => R;

type CurriedSorter<T> = CurriedFunction3<(obj: T) => any, string, T[], T[]>;

type CurriedSorterValue = {
    <T>(a: (obj: T) => any, b: string, c: T[]): T[];
}

const sortFactory: CurriedSorterValue = function (a, b, c) {
    return c;
}

Upvotes: 3

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