Why Does !New Boolean(False) Equals False in JavaScript

Why does !new Boolean(false) equals false in JavaScript?

new Boolean(false) returns an object. All objects (except document.all in browsers) are truthy.

As a result, ! of any object will always be false.

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To prove it to yourself, you can run this in your JavaScript console:

(typeof new Boolean(false)) // "object"

Also, you can use the strict equality operator === to confirm that new Boolean(false) isn’t really false:

new Boolean(false) === false // false

Incidentally, calling the Boolean function as a function—without the new—actually does return a primitive:

!Boolean(false) // true
(typeof Boolean(false)) // "boolean"

Why new Boolean(false) is true?

Says so in the docs:

Any object whose value is not undefined or null, including a Boolean
object whose value is false, evaluates to true when passed to a
conditional statement. For example, the condition in the following if
statement evaluates to true

x = new Boolean(false);
if (x) {
  // ...this code is executed
}

Directly from MDN.

Why `Number(new Boolean(false)) === 0`

As @ASDFGerte mentioned. This is because the ToNumber() method which is called by the Number() constructor will call .ToPrimitive() on the argument if an object is passed. This is why it's treated as a Boolean primitive rather than an object.

why in JavaScript's: obj = new Boolean(false) , (obj && true) is true, and (obj || false) is false?

There are two concepts which need to be considered here:

obj = new Boolean(false)

creates an Object, whose values is false. The Object itself is considered truthy, it's value (which you get with toString() or valueOf()) of course is the boolean value false.

(x||y)

returns the first truthy value (or if none are present, the last falsy value) and

(x&&y)

returns the first falsy value (or if none are present, the last truthy value).

So (obj||false) returns your Boolean Object, (obj&&true) returns the second (true) value.

The further preceeding depends on the context of your expression.

"obj && true: " + (obj && true)

demands a string context, so toString() is called on your Boolean Object, returning it's value which is false (While the object itself is truthy!).

Furthermore,

(obj && true) == true compares true == true which of course is true. However,

(obj || true) == true does a lot of type coercion §11.9.3 and compares

ToPrimitive(obj) == ToNumber(true) (§9.1 and §9.3) which results in NaN == 1 which yields false.

The results get more predictable if you use the strict equality operator §11.9.6.

new Boolean(name) always returns true

Boolean doesn't do what you think it does. It's an object, and objects are always truthy:

> typeof new Boolean(false)
"object"
> if (new Boolean(false)) console.log("It's true!");
It's true!

To convert a value to a primitive boolean type instead, a common way is to negate it twice:

const nameCheck = !!name;

You can also invoke Boolean as a function, but I'd stay away from that because it's rather confusing that it doesn't return an actual Boolean:

const nameCheck = Boolean(name);

alert(new Boolean(false)) vs console.log(new Boolean(false))

alert() displays the .toString() value of the argument that is passed to it.

The toString() value of a Boolean object is either true or false.

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