Declare interface of class only with methods signature without names

Question

Suppose I have a class with many methods, but I know for sure that their signature matches.

Is it possible to describe the interface of this class without describing the specific methods of this class in it? Like here:

interface IController {
  (input: string): number // any method without reference to its name
}

class Controller implements IController {
  method1(input: string): number { ...do something }
  method2(input: string): number { ...do something }
  ...
}

Or is it impossible?

Answer 1

The option to have an index signature (as @fk82 outlines in his answer) has the undesired consequence of forcing you to add an index signature to the class. This means that your class will be indexable by an arbitrary string, which might not be what you want.

If your goal is just to force the implementer class to only have methods with the given signature, a better option is to use a mapped type:

type IController<K extends PropertyKey> = { 
    [P in K]: (input: string) => number;
}

class Controller implements IController<keyof Controller> {
    method1(input: string): number { return input.length; }
    method2(input: string): number { return input === '' ? 0 : 1; }
}

let a = new Controller();
a['aa'] // not allowwed under no implicit any 

This has the bonus advantage of allowing the class to have some methods that do not conform to the signature if needed, but in an explicit way:

class Controller implements IController<Exclude<keyof Controller, 'special'>> {
    method1(input: string): number { return input.length; }
    method2(input: string): number { return input === '' ? 0 : 1; }
    special() { }
}