CODEWITHSUNDEEP
typescript
Anarion
Anarion

Reputation: 2534

Create an instance of generic function with a specified type

What I'm trying to achieve is something like this:

function genericFunc<T>(arg: T): T {
    return arg;
}

//this line won't work, but it's the essence of what I need
var stringFunc = genericFunc<string>;

stringFunc('myStr')

Is it possible to create an instance of a generic function with a specified type?

The only way I know is to create an interface - 

interface GenericFnInterface<T> {
    (arg: T): T;
}

let stringFunc : GenericFnInterface<string> = genericFunc;

but I would like to avoid creating too many interfaces. Is there a shorter way to achieve the same

_

UPD Another way I found is

var stringFunc: <T extends string>(arg:T) => T = genericFun;

but it's still not exactly perfect, as it creates a lot of clutter with more complex types.

Upvotes: 0

Views: 78

Answers (1)

Madara&#39;s Ghost
Madara&#39;s Ghost

Reputation: 174937

You can cast (either implicitly or explicitly): 

declare function identity<T>(arg: T): T; // implementation irrelevant

const x: (arg: string) => string = identity; // implicit cast

x('hello'); // ok
x(42); // error

or

declare function identity<T>(arg: T): T;

const x = identity as (arg: string) => string; // explicit cast

x('hello'); // ok
x(42); // error

The former is recommended, because it would catch you if you make a mistake in the cast:

const x: (arg: string) => number = identity; // implicit cast
      ~ --> Error: Type 'string' is not assignable to type 'number'.

Upvotes: 2

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