Hristo Hristo - 3 months ago 21
Javascript Question

JavaScript equality transitivity is weird

I've been reading Douglas Crockford's JavaScript: The Good Parts, and I came across this weird example that doesn't make sense to me:

'' == '0' // false
0 == '' // true
0 == '0' // true

false == undefined // false
false == null // false
null == undefined // true


The author also goes on to mention "to never use
==
and
!=
. Instead, always use
===
and
!==
". However, he doesn't explain why the above behavior is exhibited? So my question is, why are the above results as they are? Isn't transitivity considered in JavaScript?

Answer
'' == '0' // false

The left hand side is an empty string, and the right hand side is a string with one character. They are false because it is making a comparison between two un identical strings (thanks Niall).

0 == '' // true

Hence, why this one is true, because 0 is falsy and the empty string is falsy.

0 == '0' // true

This one is a bit trickier. The spec states that if the operands are a string and a number, then coerce the string to number. '0' becomes 0. Thanks smfoote.

false == undefined // false

The value undefined is special in JavaScript and is not equal to anything else except null. However, it is falsy.

false == null // false

Again, null is special. It is only equal to undefined. It is also falsy.

null == undefined // true

null and undefined are similar, but not the same. null means nothing, whilst undefined is the value for a variable not set or not existing. It would kind of make sense that their values would be considered equal.

If you want to be really confused, check this...

'\n\r\t' == 0

A string consisting only of whitespace is considered equal to 0.

Douglas Crockford makes a lot of recommendations, but you don't have to take them as gospel. :)

T.J. Crowder makes an excellent suggestion of studying the ECMAScript Language Specification to know the whole story behind these equality tests.

Further Reading?

The spec.

yolpo (on falsy values)

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