Interface type check with Typescript

Question

This question is the direct analogon to Class type check with TypeScript

I need to find out at runtime if a variable of type any implements an interface. Here\'s my

Answer 1

Because the type is unknown at run-time, I wrote code as follows to compare the unknown object, not against a type, but against an object of known type:

1. Create a sample object of the right type
2. Specify which of its elements are optional
3. Do a deep compare of your unknown object against this sample object

Here's the (interface-agnostic) code I use for the deep compare:

function assertTypeT(loaded: any, wanted: T, optional?: Set): T {
  // this is called recursively to compare each element
  function assertType(found: any, wanted: any, keyNames?: string): void {
    if (typeof wanted !== typeof found) {
      throw new Error(`assertType expected ${typeof wanted} but found ${typeof found}`);
    }
    switch (typeof wanted) {
      case "boolean":
      case "number":
      case "string":
        return; // primitive value type -- done checking
      case "object":
        break; // more to check
      case "undefined":
      case "symbol":
      case "function":
      default:
        throw new Error(`assertType does not support ${typeof wanted}`);
    }
    if (Array.isArray(wanted)) {
      if (!Array.isArray(found)) {
        throw new Error(`assertType expected an array but found ${found}`);
      }
      if (wanted.length === 1) {
        // assume we want a homogenous array with all elements the same type
        for (const element of found) {
          assertType(element, wanted[0]);
        }
      } else {
        // assume we want a tuple
        if (found.length !== wanted.length) {
          throw new Error(
            `assertType expected tuple length ${wanted.length} found ${found.length}`);
        }
        for (let i = 0; i < wanted.length; ++i) {
          assertType(found[i], wanted[i]);
        }
      }
      return;
    }
    for (const key in wanted) {
      const expectedKey = keyNames ? keyNames + "." + key : key;
      if (typeof found[key] === 'undefined') {
        if (!optional || !optional.has(expectedKey)) {
          throw new Error(`assertType expected key ${expectedKey}`);
        }
      } else {
        assertType(found[key], wanted[key], expectedKey);
      }
    }
  }

  assertType(loaded, wanted);
  return loaded as T;
}

Below is an example of how I use it.

In this example I expect the JSON contains an array of tuples, of which the second element is an instance of an interface called User (which has two optional elements).

TypeScript's type-checking will ensure that my sample object is correct, then the assertTypeT function checks that the unknown (loaded from JSON) object matches the sample object.

export function loadUsers(): Map {
  const found = require("./users.json");
  const sample: [number, User] = [
    49942,
    {
      "name": "ChrisW",
      "email": "example@example.com",
      "gravatarHash": "75bfdecf63c3495489123fe9c0b833e1",
      "profile": {
        "location": "Normandy",
        "aboutMe": "I wrote this!\n\nFurther details are to be supplied ..."
      },
      "favourites": []
    }
  ];
  const optional: Set = new Set(["profile.aboutMe", "profile.location"]);
  const loaded: [number, User][] = assertTypeT(found, [sample], optional);
  return new Map(loaded);
}

You could invoke a check like this in the implementation of a user-defined type guard.