Declaring method matches interface in TypeScript

Question

Let's say I have a class with a bunch of methods with the same signature.

class Foo {
  method1(foo: string): number { return 42; }
  method2(foo: string): number { return 99; }
}

I can define an interface for the method signature:

interface Method { (foo: string) => number; }

I would like to avoid repeating the signature over and over again for each method. I know that if I am assigning a variable I can say

const fn: Method = foo => 99;

But how can I do this when defining a method? I would like to get the equivalent of

class Foo {
  method1: Method(foo) { return 42; }
  method2: Method(foo) { return 99; }
}

but that obviously doesn't work.

Answer 1

But how can I do this when defining a method

Just like you did with a variable

class Foo {
  method1: Method = (foo) => { return 42; }
  method2: Method = (foo) => { return 99; }
}

NOTE: this does make it a member (instead of the method) but the performance implications are insignificant.