How to make (a == 1 && a == 2 && a == 3) the value true?

Address this question: https://github.com/YvetteLau/Step-By-Step/issues/9
== operator will implicit conversion.

• Object
  == operator attempts by the method valueOf and toString convert the object to its original value (a value of type string or numeric).

  const a = {
    i: 1,
    // valueOf 也可达到相同效果
    toString: function () {
      return a.i++;
    }
  }
  a == 1 && a == 2 && a == 3; // true

• Array
  for arrays of objects, toString method returns a string from the toString each element in the array () Returns the value by calling join () method connector (separated by a comma) components.

  var a = [1,2,3];
  a.join = a.shift;
  a == 1 && a == 2 && a == 3; // true

• Symbol
  when Symbol object is converted to values of the original type, toPrimitive method calls, returns the original value of the corresponding object type.

  let a = {
    [Symbol.toPrimitive]: ((i) => () => ++i) (0)
  };
  a == 1 && a == 2 && a == 3; // true

---

• Modify the get method window

  var val = 0;
  Object.defineProperty(window, 'a', {
    get: function() {
      return ++val;
    }
  });
  a == 1 && a == 2 && a == 3; // ture

• With the use of keywords
  emmm ... with keywords like rarely used

  var i = 0;
  with({
    get a() {
      return ++i;
    }
  }) {
    a == 1 && a == 2 && a == 3; // true
  }

reference:

1. Comparison operators
2. Object​.prototype​.valueOf
   
3. Establishment javascript seen from (a == 1 && a == 2 && a == 3) implicit type conversions
4. 38 questions: the following code under what circumstances would a print 1?

Reproduced in: https: //www.jianshu.com/p/a74223e37e3b