Can (a== 1 && a ==2 && a==3) ever evaluate to true

Question

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Is it ever possible that (a== 1 && a ==2 && a==3) could evaluate to true in JavaScript?

This is an interview question asked by a major tech company. It happened two weeks back, but I'm still trying to find the answer. I know we never write such code in our day-to-day job, but I'm curious.

Answer 1

Example without getters or valueOf:

a = [1,2,3];
a.join = a.shift;
console.log(a == 1 && a == 2 && a == 3);

This works because == invokes toString which calls .join for Arrays.

Another solution, using Symbol.toPrimitive which is an ES6 equivalent of toString/valueOf:

let a = {[Symbol.toPrimitive]: ((i) => () => ++i) (0)};

console.log(a == 1 && a == 2 && a == 3);

Answer 2

Okay, another hack with generators:

const value = function* () {
  let i = 0;
  while(true) yield ++i;
}();

Object.defineProperty(this, 'a', {
  get() {
    return value.next().value;
  }
});

if (a === 1 && a === 2 && a === 3) {
  console.log('yo!');
}

Answer 3

Yes, it is possible!

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Answer 4

Honestly though, whether there is a way for it to evaluate to true or not (and as others have shown, there are multiple ways), the answer I'd be looking for, speaking as someone who has conducted hundreds of interviews, would be something along the lines of:

"Well, maybe yes under some weird set of circumstances that aren't immediately obvious to me... but if I encountered this in real code then I would use common debugging techniques to figure out how and why it was doing what it was doing and then immediately refactor the code to avoid that situation... but more importantly: I would absolutely NEVER write that code in the first place because that is the very definition of convoluted code, and I strive to never write convoluted code".

I guess some interviewers would take offense to having what is obviously meant to be a very tricky question called out, but I don't mind developers who have an opinion, especially when they can back it up with reasoned thought and can dovetail my question into a meaningful statement about themselves.

Answer 5

This is an inverted version of @Jeff's answer* where a hidden character (U+115F, U+1160 or U+3164) is used to create variables that look like 1, 2 and 3.

var  a = 1;
var ﾠ1 = a;
var ﾠ2 = a;
var ﾠ3 = a;
console.log( a ==ﾠ1 && a ==ﾠ2 && a ==ﾠ3 );

* That answer can be simplified by using zero width non-joiner (U+200C) and zero width joiner (U+200D). Both of these characters are allowed inside identifiers but not at the beginning:

var a = 1;
var a‌ = 2;
var a‍ = 3;
console.log(a == 1 && a‌ == 2 && a‍ == 3);

/****
var a = 1;
var a\u200c = 2;
var a\u200d = 3;
console.log(a == 1 && a\u200c == 2 && a\u200d == 3);
****/

Other tricks are possible using the same idea e.g. by using Unicode variation selectors to create variables that look exactly alike (a︀ = 1; a︁ = 2; a︀ == 1 && a︁ == 2; // true).

Answer 6

It can be accomplished using the following in the global scope. For nodejs use global instead of window in the code below.

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

This answer abuses the implicit variables provided by the global scope in the execution context by defining a getter to retrieve the variable.