Interviewer: Can (a==1 && a==2 && a==3) Ever Evaluate to ‘true’ in JavaScript?

 

The expression `(a == 1 && a == 2 && a == 3)` evaluates to `true` seems counterintuitive at first glance because `a` is expected to have a single value, and it's unlikely for it to simultaneously equal `1`, `2`, and `3`. However, JavaScript allows for customizing object property access using getters, which can lead to unexpected behavior and manipulations of how values are retrieved.

Here's an example of how you can make `a` evaluate to `true` in that expression:

let a = {

  value: 1,

  valueOf: function() {

    return this.value++;

  }

};

console.log(a == 1 && a == 2 && a == 3); // Output: true

Explanation:

- By defining a custom `valueOf` method within the object `a`, we've overridden how the value of `a` is retrieved.

- Every time a is compared, the `valueOf` method is invoked, returning `this.value++`. This increments the `value` property of `a` each time it's compared.

- So, when the expression (a == 1 && a == 2 && a == 3) is evaluated, a behaves as if it has different values (1, 2, and 3) successively in each comparison due to the valueOf method being called.

This code manipulates the equality checks by dynamically changing the value of a on comparison, allowing it to pass multiple equality checks within a single expression.

This behavior, while technically possible in JavaScript, is generally considered a code smell and can make the codebase hard to understand and maintain. It's essential to write clear and understandable code for better readability and maintainability.