Defining the interface for a function using function properties

Question

As you may know, functions in JavaScript can have properties as any object. For example (taken from the excellent JavaScript: The Definitive Guide, 6th ed, p. 178) computes a factorial using the function as memoization array:

function factorial(n: number): number {
  if (isFinite(n) && n > 0 && n == Math.round(n)) {
    if (!(n in factorial))
      factorial[n] = n * factorial(n - 1);
    return factorial[n];
  }
  else
    return NaN;
}
factorial[1] = 1;

I tried defining the following interface:

interface Factorial {
  (n: number) : number;
  [ index: number ]: number;
}

But the compiler is telling me that Type '(n: number) => number' is not assignable to type 'Factorial'. Index signature is missing in type '(n: number) => number'. I can't do the obvious thing and just define private index: number; inside the function, I'm stumped.

Saravana · Accepted Answer · 2017-06-21 15:56:19Z

1

What you have is an example of hybrid types. You have to use type assertion to make sure the function complies with the interface:

let factorial = function (n: number): number {
    if (isFinite(n) && n > 0 && n == Math.round(n)) {
        if (!(n in factorial))
            factorial[n] = n * factorial(n - 1);
        return factorial[n];
    }
    else
        return NaN;
} as Factorial;

factorial[1] = 1;

answered Jun 21, 2017 at 15:56

Saravana

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1 Comment

that_individual Over a year ago

Thought there was a better way, but I guess this is it. It does the trick. Thanks

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