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?

Answer 1

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.