JavaScript equality transitivity is weird

Question

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\'                


        

Answer 1

'' == '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)