Explaining loose-equality in javascript

Alejandro Garcia Serna — Fri, 03 May 2019 04:09:50 +0000

One of the most "strange" behaviors of javascript is how the == equality operator works. You can see pieces of code like this:

  42 == '42' // true

On the internet, you'll find one common explanation about this, and is that == operator compares only the value in both sides of the equality without looking the type. This is a correct explanation but is not very accurate.

One more accurate explanation is that == reduces both sides of the equality to their number representation and there, it makes the comparison. Consider the following:

  0 == false // true

Zero is already a number, but what happens with false?
If you convert false to a number you'll get this

  Number(false) // 0

Then, the final comparison will be this

 0 == 0 // true

Having this in mind, check this out:

 [] == ![] // true

The comparison above, follows the same rule that I've mentioned before. the empty array [] if is converted to a number, gives as result:

 Number([]) // 0

On the other side of the comparison, we have ![] as we have the ! operator, this one changes the boolean value of the empty array. The boolean representation of [] is:

 Boolean([]) // true

And as we have the ! operator the result is false, and as we saw above, the number representation of false is 0.

As final comparison we are going to have:

 0 == 0 // true

I really hope that this blog post help you to understand this behavior of javascript, the feedback is welcome and if you like this content I would be very happy if you share it.
Thanks for reading.

DEV Community: Alejandro Garcia Serna

Explaining loose-equality in javascript