Question 245470 in Javascript Question posted in
A very good W3school tutorial can be found here.

Javascript – Make polyfills immutable

For example:

Object.defineProperty(Promise.prototype, 'onFinally', {
  get: () => {},
  writable: false,
});

or

Object.freeze(Promise.prototype);

These examples aren’t work, is there a working way?

Tags:

2

Answers


  1. 0

    You can call Object.freeze() on the String.prototype, but if you try to redefine a method, it will silently be ignored.

    Note that freezing an Object’s prototype is ill-advised.

    function padBoth(targetLength, padString) {
  const delta = Math.floor((targetLength - this.length) / 2);
  return delta > 0
    ? this
      .padStart(this.length + delta, padString)
      .padEnd(targetLength, padString)
    : this
}

String.prototype.padBoth = padBoth;
console.log('Hello World'.padBoth(20, '-'));

// Without this line, the try-catch (below) will throw
Object.freeze(String.prototype);

// No error, but does not reassign
String.prototype.padBoth = undefined;
try {
  console.log('Hello World'.padBoth(20, '-')); // Works, if frozen
} catch (e) {
  console.log(e.message);
}
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  2. 0

    Actually the example you provided works:

    Object.defineProperty(Promise.prototype, 'onFinally', {
    get: () => { console.log("Retrieved.") }
});

Promise.prototype.onFinally

Object.freeze(Promise.prototype);

Promise.prototype.onFinally

Object.defineProperty(Promise.prototype, 'onFinallySecond', {
    get: () => { console.log("Retrieved.") }
});

    TypeError: Cannot define property onFinallySecond, object is not extensible

    You just have to call it in the correct order.

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