how to create typescript definitions for function that extend objects

Question

there are function in 3rd party libraries that extends objects to add to them more functions. e.g. seamless-immutable

I want to create definition file for such a library. I thought something in the line of

interface ImmutableMethods<T> extends T{
    set?<U>(key: string | number, value: any): T | U;
    setIn?<U>(keys: Array<string>, value: any): T | U;
    ...
}
function Immutable<T>(obj: T, options?): ImmutableMethods<T>

Then I will be able to declare my own types as:

interface Type1 {prop1, prop2}
interface Type2 {propOfType1: ImmutableMethods<Type1>}
var immutableType2:Type2 = Immutable(...)

And then I'll have the Immutable's function on all my objects.

But the problem is that the line interface ImmutableMethods<T> extends T gives me an error "An interface may only extend a class or another interface"

Is there a way to declare that T is an interface, or a whole other way to get this extending behavior?

I don't see why you want to use extends T on your interface. Can't you just extend from Immutable<T> — Icepickle
– Icepickle, Commented Feb 23, 2016 at 15:22
I don't want to couple my business logic interface with implementation details of the library I'm working with. Another thing is that there is a need to pass arrays and mapping so instead of just writing x: ImmutableMethods<Array<MyInterface>>, I need to somehow create a new kind of Array object. — ziv
– ziv, Commented Feb 25, 2016 at 6:47
FYI, I posted an issue on the seamless-immutable repo regarding typings: github.com/rtfeldman/seamless-immutable/issues/108 — Jonas Kello
– Jonas Kello, Commented Mar 7, 2016 at 12:07

Paarth · Accepted Answer · 2016-04-21 07:30:41Z

2

You can use intersection types to attain this behavior. Define an interface for the extra functions and define a type that is T & Interface. My modifications to your code:

interface ImmutableMethods<T> {
    set?<U>(key: string | number, value: any): T | U;
    setIn?<U>(keys: Array<string>, value: any): T | U;
}
type ImmutableExtended<T> = ImmutableMethods<T> & T;

interface Type1 { prop1: string, prop2: string }
interface Type2 { propOfType1: ImmutableExtended<Type1> }
var immutableType2: Type2 = ...;

The ImmutableMethods<T> interface defines the functions involved. The ImmutableExtended<T> type is the interface we actually want to use for the extended objects, though of course you can name these interfaces whatever you like.

answered Apr 21, 2016 at 7:30

Paarth

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