Using TypeScript, is there a way to define an object of which each key is the same type, while prese-CodePudding

I have a type for a "helper" function which receives a context object as first parameter:

type helper = ((ctx: Context, ...params: any[]) => any);

Then I want to create an object of which each value is of the helper type:

const lib = {
  path: (ctx) => ctx.getPath(),
  array: (ctx, ...params) => new MyArray(ctx, params),
};

Is there a way to define the type of lib in this example so that each contained function already knows its of the helper type and doesn't need to define the type of ctx over and over? So far, I know you can declare it like this:

const lib: {[key: string]: helper} = {
  path: (ctx) => ctx.getPath(),
  array: (ctx, ...params) => new MyArray(ctx, params),
};

But then the types would lose the information of lib containing the keys "path" or "array" since it can contain any key. Another alternative would be to define the type of every function like so:

const lib = {
  path: (ctx: Context) => ctx.getPath(),
  array: (ctx: Context, ...params: any[]) => new MyArray(ctx, params),
};

But that feels inefficient since we already know that all values of the lib object are of the same type. Is there a way to declare its type and get the best of both alternatives?

CodePudding user response：

You're looking for the new

Side note: In the above and linked playground I've capitalized Helper because it's the standard convention for non-primitive types to start with a capital letter.