Why if([]) is validated while [] == false in javascript?

Question

if([] == false) alert(\'empty array is false\');
alert(+[]) // alert 0
if([]) alert(\'empty array is true\');

They both will run the alert

Answer 1

It's because of type coercion of the == (equality) operator.

An empty array is considered truthy (just like an empty object), thus the second alert is called.

However, if you use ([] == false), your array is coerced to its string representation* which is "" which then is considered as a falsy value, which makes the condition true thus triggering the first alert too.

If you want to avoid type coercion, you have to use the === (identity) operator which is the preferred and by the famous Douglas Crockford promoted way to compare in javascript.

You can read more on that matter in this exhaustive answer.

*^{(Object.prototype.toString is called on it)}

EDIT: fun with JS-comparison:

NaN == false // false
NaN == true  // also false
NaN == NaN   // false

if(NaN)      // false
if(!NaN)     // true

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

This shows you the real "power" of Comparison with == due to the strange rules mentioned in bfavarettos answer.