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Typescript creating own types with functions

By : J Bolt
Date : July 30 2020, 01:00 AM
Hope this helps
The Problem is that i could not find where and how the function bodys gets defined.
code :
class Byte {
    constructor(public value: number) {
    yourMethodHere() {
        // ...
// Adding it to the type
interface Number {
    toHex(): string;
// Adding the implementation
Object.defineProperty(Number.prototype, "toHex", {
    value() {
        return this.toString(16);
    enumerable: false, // This is the default, but just for emphasis...
    writable: true,
    configurable: true

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TypeScript: types for variadic functions

By : rinku
Date : March 29 2020, 07:55 AM
I wish did fix the issue. , In TypeScript you can use "..." to achive the above pattern:
code :
function sum (...numbers: number[]) {
  var sum = 0;
  for (var i = 0; i <  numbers.length; i++) {
    sum += numbers[i];
  return sum;

Typescript: `this` types and generic functions

By : Jorge Marcelo Peña A
Date : March 29 2020, 07:55 AM
Any of those help The last version from the question was almost there. I used wrong syntax to the constraint on the generic type. The final version looks like this
code :
interface Addable<T> {
    add: (this: Addable<T>, other: Addable<T>) => Addable<T>;
class Point implements Addable<Point> {
    constructor(public x: number, public y: number) { }
    add(other: Point) {
        return new Point(this.x + other.x, this.y + other.y);

class Num implements Addable<Num> {
    constructor(public number: number) { }
    add(other: Num) {
        return new Num(this.number + other.number);
const add: <T extends Addable<T>>(a1: T, a2: T) => T = function (a1, a2) {
    return a1.add(a2);

var p1 = new Point(1, 1);
var p2 = new Point(2, 2);
var num1 = new Num(10);

const added = p1.add(p2);
const added2 = add(p1, p2);

const wrongAdded = p1.add(num1);  // Fails

const wrongAdded2 = add(p1, num1);  // Fails

Mapped Types in Typescript with functions

By : user2337169
Date : March 29 2020, 07:55 AM
Any of those help Your problem is actually quite simple. inferredSelectorCreator takes in a rest parameter of type T, but when you call you call inferredSelectorCreator with the whole array, without spreading it (inferredSelectorCreator(selectors)) this means T will be inferred to [[Selector, Selector]] instead of [Selector, Selector].
If you use a spread you get the expected result:
code :
const bazz = inferredSelectorCreator(...selectors)

Enforce that two functions take the same parameter types in TypeScript

By : user3732650
Date : March 29 2020, 07:55 AM
seems to work fine You can add another type parameters for the parameters and use tuples in rest parameters to spread it to the function signature. This will ensure the parameter types of the two functions are the same (or at least compatible)
code :
type Func<TParams extends any[], TResult> = (...args: TParams) => TResult;

function foo<TParams extends any[], TResult>(f: Func<TParams, TResult>, g: Func<TParams, TResult>): TResult {
    return null!

function f1(s: string, i: number): boolean { return true; }
function g1(s: string, i: number): number { return 0; }
foo(f1, g1);

function g2(b: boolean): boolean { return false; }
foo(f1, g2);

How to allow partial TypeScript types when mocking functions - Jest, TypeScript

By : yash
Date : March 29 2020, 07:55 AM
This might help you I have a function which I would like to mock for testing purposes in TypeScript. In my tests, all I care about are the json and the status. However, when using Jest's jest.spyOn the type of my mocked function is set to return a http Response type. This is awkward as it means I have to manually go and implement a bunch of functions and properties that are irrelevant and arbitrary. , You can use as...
code :
export function mockApi(json: object, status: number): void {
  ).mockImplementation(() =>
      json: () => Promise.resolve(json),
    } as http.Response), // <-- here
type X {
  a: number
  b: number

const x = { a: 2 } as X // OK
const y = { a: 3, c: 2 } as X // NOT OK, because c does not exist in X
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