I’m trying to get a clear idea about how the strict equality operator functions in JavaScript. I have a concept in mind, but I’m not confident it’s completely right.
Is the following true: x === y means (typeof x === typeof y) && (x == y) always?
For instance:
let firstValue = 42;
let secondValue = "42";
let thirdValue = 42;
// Will these comparisons yield the same result?
console.log(firstValue === thirdValue); // true
console.log((typeof firstValue === typeof thirdValue) && (firstValue == thirdValue)); // true?
console.log(firstValue === secondValue); // false
console.log((typeof firstValue === typeof secondValue) && (firstValue == secondValue)); // false?
Can anyone verify if this reasoning is valid in all scenarios or provide a case where it doesn’t apply? I want to ensure I grasp the key differences between strict and loose equality.
Actually there’s a more fundamental issue with your approach that the previous answer touched on but didn’t fully explain. The triple equals operator follows a specific algorithm defined in the ECMAScript specification that doesn’t involve the loose equality operator at all. When you use x === y, JavaScript first checks if the types are identical. If they’re not, it immediately returns false without any further processing. If the types match, it then performs a strict value comparison based on the specific type rules. The problem with your formula is that x == y can return true even when the values are fundamentally different due to type coercion. For example, null == undefined returns true, but null === undefined returns false. Your formula would suggest (typeof null === typeof undefined) && (null == undefined), but typeof null gives “object” while typeof undefined gives “undefined”, so the first part fails anyway. Another case: NaN === NaN is always false, but NaN == NaN is also false. However, the reasoning behind these results differs completely. The strict equality has special rules for NaN that don’t rely on loose equality logic.
Your understanding has the right general concept but it’s not technically accurate in all cases. The triple equals operator doesn’t actually perform a loose equality check as part of its process. Instead, it directly compares both type and value without any type coercion whatsoever.
The key difference is that === will immediately return false if the types don’t match, without proceeding to any value comparison. Meanwhile, your formula (typeof x === typeof y) && (x == y) could potentially give different results because the == operator performs type coercion.
Consider this edge case: 0 === false returns false, but (typeof 0 === typeof false) && (0 == false) would actually be false && true, which is false. However, the reasoning process is different. The == part evaluates to true because JavaScript coerces the boolean to a number, but === never gets that far since the types differ from the start.
So while your examples work correctly, the fundamental mechanism isn’t quite right. Think of === as a single-step operation that checks both type and value simultaneously, rather than a two-step process involving loose equality.
your formula breaks down with object comparisons too. try {} === {} vs (typeof {} === typeof {}) && ({} == {}) - both give false but for completly different reasons. triple equals checks reference equality for objects while loose equality does weird coercion stuff first.