JavaScript !!1==”1″ True and !!2==”2″ False – Explanation

javascript

As the title states, why does:

> !!1=="1"

equal

True

and

> !!2=="2"

equal:

False

Likewise, why does > "1"==true equal true and > "2"==true equal false

I'm baffled. Are these just bugs in JS or what's going on here?

Best Answer

As per the Operator precedence rules, logical ! has higher priority over ==. So, in both the cases, !! is evaluated first.

Note: Truthiness of various objects have been explained in this answer of mine.

First Case

!!1 == "1"

!1 will be evaluated to false, since 1 is considered Truthy. Negating again we get true. So the expression becomes

true == "1"

Now, the coercion rules kick in as you have used == operator, which evaluates as per the The Abstract Equality Comparison Algorithm defined in ECMAScript 5.1 Specification,

6. If Type(x) is Boolean, return the result of the comparison ToNumber(x) == y.

So, true will be converted to a number, which is 1 as per ToNumber algorithm for Boolean values. Now the expression becomes

1 == "1"

Now,

4. If Type(x) is Number and Type(y) is String, return the result of the comparison x == ToNumber(y).

So, "1" will be converted to a number and that will give 1, as per the ToNumber algorithm. That is why it shows true in the first case.

Second Case

The same rules are applied here.

!!2 == "2"

becomes

true == "2"

then

1 == "2"

which becomes 

1 == 2

which is not true, that is why the second case prints false.

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