In JavaScript, why is “0” equal to false, but when

Question

The following shows that "0" is false in Javascript:

>>> "0" == false
true

>>> false == "0"
true

So why does the following print "ha"?

>>> if ("0") console.log("ha")
ha

Answer 1

Tables displaying the issue:

and ==

Moral of the story use ===

table generation credit: https://github.com/dorey/JavaScript-Equality-Table

Answer 2

The "if" expression tests for truthiness, while the double-equal tests for type-independent equivalency. A string is always truthy, as others here have pointed out. If the double-equal were testing both of its operands for truthiness and then comparing the results, then you'd get the outcome you were intuitively assuming, i.e. ("0" == true) === true. As Doug Crockford says in his excellent JavaScript: the Good Parts, "the rules by which [== coerces the types of its operands] are complicated and unmemorable.... The lack of transitivity is alarming." It suffices to say that one of the operands is type-coerced to match the other, and that "0" ends up being interpreted as a numeric zero, which is in turn equivalent to false when coerced to boolean (or false is equivalent to zero when coerced to a number).

Answer 3

It is all because of the ECMA specs ... "0" == false because of the rules specified here http://ecma262-5.com/ELS5_HTML.htm#Section_11.9.3 ...And if ('0') evaluates to true because of the rules specified here http://ecma262-5.com/ELS5_HTML.htm#Section_12.5

Answer 4

// I usually do this:

x = "0" ;

if (!!+x) console.log('I am true');
else      console.log('I am false');

// Essentially converting string to integer and then boolean.