JavaScript type casting

Question

Consider empty JavaScript array:

var a = [];
alert(a == false); // shows true
alert(!a); // shows false!

How to explain this? What are the rules?

Solution 1

From http://forums.whirlpool.net.au/archive/966449:

a == false:

In this case, the type of the left-hand side is object, the type of the right-hand side is boolean. Javascript first converts the boolean to a number, yielding 0. Then it converts the object to a "primitive", yielding the empty string. Next it compares the empty string to 0. The empty string is converted to a number, yielding 0, which is numerically equal to the 0 on the right-hand side, so the result of the entire expression is true.

See §11.9.3 of the ECMAScript spec for all the gory details.

(!a):

In this case Javascript converts the object to the boolean true, then inverts it, resulting in false.

Solution 2

The ! operator checks whether its operand is "falsy".

The following are true:

• !false
• !0
• !null
• !NaN
• !undefined
• !""

---

The == operator checks for loose equality, which has nothing to do with falsiness.

Specifically, a == b will convert to operands to numbers, then compare the numbers.
Strings containing numbers convert to the numbers that they contain; booleans convert to 0 and 1.
Objects are converted by calling valueOf, if defined.

Thus, all of the following are true:

• "1" == 1
• "0" == false
• "1" == true
• "2" != true
• "2" != false
• ({ valueOf:function() { return 2; } }) == 2
• ({ valueOf:function() { return 1; } }) == true

Solution 3

The == operator when one of the operands if Boolean, type-converts the other to Number.

[] == 0;

Is equivalent to:

0 == 0;

You can see the complete details of The Abstract Equality Comparison Algorithm on the specification.

As you can see, an empty array object, when converted to Number, produces 0:

+[]; // 0
Number(0);

This is really because its toString method produces an empty string, for example:

[].toString(); // ""

+""; // 0
Number(""); // 0

Solution 4

When comparing an object to a primitive value via the == operator, the object coerces into an primitive value itself (number or string). In this case [] coerces into 0, then false coerces into 0:

[] == false
0 == false
0 == 0

which is true.

The ! operator coerces into boolean and then inverts the value. [] into boolean is true (like with any object). Then invert to become false

![]
!true
false

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Comments

[qwertymk]

How are ("2" != true), ("2" != false) both true?

[SLaks]

@qwerty: The same way that 2 != 3 and 2 != 4 are both true

[qwertymk]

Oh because false == Number(0), true == Number(1) . Got it thanks

[JKirchartz]

here's a direct link to the afore-mentioned ECMA spec: ecma-international.org/ecma-262/5.1/#sec-11.9.3