What is the purpose of new Boolean() in Javascript? - javascript

What is the use of:
var flag = new Boolean(false);
compared to:
var flag = false;
When would you actually use new Boolean?

The global function Boolean() can be used for type casting when called without new, eg
var foo = Boolean(bar); // equivalent to `var foo = !!bar`
When called with new, a wrapper object will be created additionally, which means that you can assign arbitrary properties to the object:
var foo = new Boolean(bar); // equivalent to `var foo = Object(Boolean(bar));`
foo.baz = 'quux';
alert(foo.baz);
This is not possible with primitive values as primitives can't hold properties:
var foo = true;
foo.baz = 'quux';
alert(foo.baz); // `foo.baz` is `undefined`
Assigning a property to a primitive doesn't produce an error because of auto-boxing, ie
foo.baz = 'quux';
will be interpreted as
// create and immediately discard a wrapper object:
(new Boolean(foo)).baz = 'quux';
To get the primitive value back, you'll have to invoke the valueOf() method. This is needed if you want to actually use the wrapped value, because objects always evaluate to true in boolean contexts - even if the wrapped value is false.
I've never come across a useful application of being able to assign properties to booleans, but boxing might be useful in cases where a reference to a primitive value is needed.

While others mentioned the theory, let me talk about the practical part:
Because Boolean objects (as objects in general) are always truthy, it is considered bad practice to use them. In many years of JS programming, I have never used them, and I can't remember seeing Booleans in other peoples' code either. Not even once.
Using primitive values will avoid confusion and will make your code a little bit shorter.
If you ever need a bool wrapped in an object, you might as well use an Object object like so:
foo = { value: false };
Also, calling the Boolean() constructor as a function (as in foo = Boolean(bar)) has the same effect as explicit typecasting using !!, and the latter is generally preferred over the former.

Before the above question first the Boolean function, Boolean ()
Boolean(10 > 4) // return true
Boolean(4 > 9) // return false
Next: everything with real value return true. E.g
100
-4
4.4
"hello"
"false" // note even the string value false return true.
everthing without real value return false E.g
NaN
var x = 10 / "H"; // Boolean(x); return false.
undefined
""
0
-0
false
null
Now the Boolean object is an object wrapper for a boolean value. The value passed as the first parameter is converted to a boolean value, if necessary. If value is omitted or is 0, -0, null, false, NaN, undefined, or the empty string (""), the object has an initial value of false. All other values, including any object or the string "false", create an object with an initial value of true.
This allows very powerful tricks.

Interesting question:
You use new Boolean to create a boolean object. There can be many scenarios but I have discussed below one scenario.
Suppose you want a comparison in your code where you want to match string value and its datatype and it has to bool (true/false) then you will use new boolean instead of assigning simple false value.
var flag = false;
var flag2 = new Boolean (false);
alert(typeof flag); //boolean object
alert(typeof flag2); //simple object
if (flag === flag2){
alert("Value and datatype match");
}
else{
alert("Value and datatype do not match");
}

Related

Boolean coercion fails

I am wondering why the Boolean coercion fails in this case:
!!(new Boolean(false)) === true
Although:
(new Boolean(false).valueOf()) === false
Mozilla says:
Booleans are returned as-is.
I am wondering what "as-is" means in the context of a coercion. I thought "coercion" means "convert anything to a primitive boolean". How is it possible that something which is meant to be false gets coerced to true?
BTW: Consequently this fails too:
Boolean(new Boolean(false)) === true
It seems to me, that the Boolean class itself is an error. Maybe I have to use the following code:
if (arg instanceof Boolean)
throw new Error("Stop doing nonsense")
Or maybe this:
function coerce_boolean_correctly (arg) {
if (arg instanceof Boolean)
return coerce_boolean_correctly(arg.valueOf())
return !!arg
}
There is a huge difference between a Boolean object and a Boolean primitive. The MDN page you referenced, actually warns about this at the very outset:
Do not confuse the primitive Boolean values true and false with the true and false values of the Boolean object.
Any object, including a Boolean object whose value is false, evaluates to true when passed to a conditional statement.
And this (what I marked in bold) is exactly what happens in your code: it creates a new Boolean(false) and that will coerce to true -- in other words it is a truthy value. In your code you have explicitly converted it to a boolean primitive, by applying ! to it twice. In either case (implicit coercion or explicit conversion) new Boolean(false) is truthy. Fact is that all objects are considered truthy (when coerced to boolean, they evaluate to true).
The article continues with:
Do not use the Boolean() constructor with new to convert a non-boolean value to a boolean value — use Boolean as a function or a double NOT instead.
This suggests that your code should be modified to drop the use of new, and call Boolean as a plain function, not as constructor:
!!(Boolean(false)) === false
When Boolean is called as plain function, it returns a primitive boolean (false or true). But anything that is called as constructor, even Boolean, will return an object. And objects are truthy.
When in the context of coercion MDN states "Booleans are returned as-is." they refer to boolean primitives. (Boolean) objects are covered by the last bullet point in the same list: "All objects become true". The latter in includes Boolean objects.
A Boolean object has a valueOf method which returns a boolean primitive, and so it returns what you would intuitively expect.
The MDN article rightly says:
Warning: You should rarely find yourself using Boolean as a constructor.
Don't use new Boolean, but Boolean.
If for some reason you have a Boolean object, really ask yourself why you have that object in the first place. Tackle the code that created that object and use a primitive boolean from the outset.
Objects are truthy
Some are surprised in a similar way that the following holds:
if (!!new String("")) console.log("An empty string is truthy!?");
if (!!new Number(0)) console.log("The number 0 is truthy!?");
if (!!new Number(NaN)) console.log("NaN is truthy!?");
if (!!new Object(null)) console.log("null is truthy!?");
if (!!new Object(undefined)) console.log("Undefined is truthy!?");
It is the same principle: objects are truthy. Always.* No matter what their constructor is named. No matter what value was passed to that constructor.
* document.all is an abhorrent, but accepted, violation of this rule.
See also
What's the point of the Boolean object?
What is the purpose of new Boolean() in Javascript?

How do I assign a boolean object to a value?

I'm looking at what the Boolean object works. It can't be used as reference value transfer. Because the value can't be changed. What's the use of using the new operator? The problem can be solved by creating a new boolean object myself. But that doesn't answer the question, why doesn't there be a set function in the boolean object? Creating a boolean object means nothing to me. Because you can't do anything. Is there a solution other than creating a new object? and what does the boolean object do?
let bool=true;
let boolObj=new Boolean(true);
let ref=boolObj;
//set booObj false
console.log(ref.valueOf());//expected out:false
Is there a solution other than creating a new object?
If the problem is that you want an object with mutable boolean state, then yes, that's the solution:
const boolObj = {flag: true, valueOf() { return this.flag; }};
boolObj.flag = false;
console.log(boolObj.valueOf());
Note that that uses boolObj.flag = false;, not boolObj = false;. The latter would store false in boolObj, not modify the state of the object.
What's the use of using the new operator?
There is almost never any reason to create a Boolean object. Boolean objects, like all the primitive wrapper objects, are immutable. Their primary purpose is to provide a specification mechanism by which "methods" can be called on primitives:
const b = false;
console.log(b.toString()); // "false"
b is a primitive, so how can it have a method? The answer is: It can't. But when the JavaScript engine sees b.toString, it "promotes" the primitive to the equivalent object and then gets the method from that equivalent object. (Theoretically; obviously, engines optimize that object creation away when they can.) This doesn't mean much with the built-in methods, but you can add methods. Before ES5's strict mode, that meant that there had to be an object representation of the primitive:
Object.defineProperty(Boolean.prototype, "example", {
value() {
console.log("typeof this = " + typeof this);
return !this;
},
configurable: true,
writable: true
});
false.example();
That was necessary because before ES5's strict mode, this always had to be of type "object" (null or a reference to an object), it couldn't be a primitive. As of ES5's strict mode, this is allowed to be a primitive:
"use strict";
Object.defineProperty(Boolean.prototype, "example", {
value() {
console.log("typeof this = " + typeof this);
return !this;
},
configurable: true,
writable: true
});
false.example();
I said "almost never" above because there is one use case for Boolean objects, but it's not a very good one: Tristate flags. You can use null for the indeterminate state, new Boolean(true) for the true state, and new Boolean(false) for the false state. Again, though, it's not a great use case. :-)

Why is 'true instanceof Boolean' equal to false in JavaScript [duplicate]

This question already has answers here:
Why does instanceof return false for some literals?
(10 answers)
Closed 7 years ago.
The following indicates the expression "true instanceof Boolean" evaluates to false. Why would this expression evaluate to false?
$(document).ready(function() {
var $result = $('#result');
if(true instanceof Boolean) {
$result.append('I\'m a Boolean!');
} else {
$result.append('I\'m something other than a Boolean!');
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="result"></div>
The value true is a boolean primitive — that's "boolean" with a lower-case "b". It is not an object.
Similarly, "hello" is not an instance of String, because a string primitive is not an object either.
You can test for primitive types with the typeof operator.
The distinction between primitive values and object values can be a little confusing in JavaScript because the language implicitly wraps primitives in appropriate object wrappers when primitives are used like object references. That's why you can write
var length = "hello world".length;
The string is implicitly converted to a String instance for that . operation to work. You can even do that with a boolean:
var trueString = true.toString();
It happens with numbers too, though the syntax gets in the way sometimes:
var failure = 2.toString(); // an error
var success = (2).toString(); // works
The Boolean object is an object wrapper for a boolean value.
So the boolean value and the wrapper object are not the same thing.
Also, boolean objects are verboten. In other words it isn't good style to ever use a Boolean object.

Javascript complicated valueOf method [duplicate]

This question already has answers here:
Object.prototype.valueOf() method
(2 answers)
Closed 9 years ago.
What does ({}).valueOf.call(myvar) do?
it converts any value to an object (an object remains unchanged, a primitive is converted to an instance of a wrapper type).
My question is how?Can someone give The longer answer how this is done behind the scene.Since valueOf() method is meant to return primitive values not object .
console.log{name:"sameer"}.valueOf() //returns an object but cant be displayed since toString() method will be called by js so [object Object] gets displayed which is a string ,how to display the exact return value from valueOf() method .Is there a way?
Hello again! Once more, we face the mighty opponent. Before we begin, let's dispel one false thought:
valueOf() method is meant to return primitive values not object .
Not accurate. valueOf returns an object if a primitive value was passed to it. If you do valueOf(object), you'd get the same object: valueOf(object) === object. You can trivially see that:
var obj = {};
obj.valueOf() === obj; //true
Now, for the more interesting question: How is valueOf defined? Let's look at the ES5 specification along with the v8 and spidermonkey sources.
valueOf (spec, v8, spidermonkey):
function ObjectValueOf() {
return ToObject(this);
}
As we can see, it simply returns ToObject, as defined in the spec. The rabbit hole emerges.
ToObject (spec, v8, spidermonkey)
function ToObject(x) {
if (IS_STRING(x)) return new $String(x);
if (IS_SYMBOL(x)) return new $Symbol(x);
if (IS_NUMBER(x)) return new $Number(x);
if (IS_BOOLEAN(x)) return new $Boolean(x);
if (IS_NULL_OR_UNDEFINED(x) && !IS_UNDETECTABLE(x)) {
throw %MakeTypeError('null_to_object', []);
}
return x;
}
Jackpot. We can see the entire flow here. If it's a string, number, boolean, etc return a wrapper ($String and $Boolean and the likes represent the actual String or Number; see here); if it's an invalid argument, throw an error; otherwise, return the argument.
The spidermonkey source for that one goes deeper down the rabbit hole. It defines ToObject as such:
JS_ALWAYS_INLINE JSObject *
ToObject(JSContext *cx, HandleValue vp)
{
if (vp.isObject())
return &vp.toObject();
return ToObjectSlow(cx, vp, false);
}
So if it's not an Object, call ToObjectSlow. Buckle up Alice, there'll be C++. We need to take a look at what ToObejctSlow does:
JSObject *
js::ToObjectSlow(JSContext *cx, HandleValue val, bool reportScanStack)
{
if (val.isNullOrUndefined()) {
...error throwing magic here...
return NULL;
}
return PrimitiveToObject(cx, val);
}
More indirection after looking whether the argument was null or undefined. The finale is here:
JSObject *
PrimitiveToObject(JSContext *cx, const Value &v)
{
if (v.isString()) {
Rooted<JSString*> str(cx, v.toString());
return StringObject::create(cx, str);
}
if (v.isNumber())
return NumberObject::create(cx, v.toNumber());
JS_ASSERT(v.isBoolean());
return BooleanObject::create(cx, v.toBoolean());
}
Pretty much the same as the v8 version, only with different taxonomy.
Now, as I said before, I think your question has more to do with the medium of representing the object you see. Firebug and chrome's devtools are more than apt at displaying an object. However, if you try to alert it, you'll see the unfortunate [object Object], because that's what ({}).toString() gives you (since it gives out a string of the form [object InternalClassName], again, as we've seen before).
As a bonus, try console.dir({foo : 'bar'})
To answer your first question
JavaScript has two main variable category types, primitives and Objects. You will often hear this, in JS everything is an Object. That is not entirely accurate. There are also primitive data types, which do nothing but hold values.
They have no methods and they are not instances of a wrapper type. So before you can call any method on them, they need to be converted to an object of the wrapper type. In JavaScript this conversion is automatic and it is called auto-boxing.
Allow me to demonstrate:
var firstString = "Test";
typeof firstString == "string"; // true
var secondString = new String("Test");
typeof secondString == "string"; // false
secondString.prototype.toString.call// [object String];
Notice what happens. There are actually two types above. One is string and the other one is [object String]. This tells you two things: secondString instanceof String is true. That is a wrapper type. Inside the core language you are seeing that String inherits from Object.
But the first string is just a memory reference, nothing more. When you call methods like firstString.replace(), firstString is automatically converted to its wrapper type. This is autoboxing.
The above behaviour stands for the following pairs:
Number autoboxing
var x = 5; var y = new Number(5);,
Boolean autoboxing
var x = false; var y = new Boolean(false);
RegExp autoboxing
var x = new RegExp("etc"); var y = /etc/;
Object.prototype.valueOf
The valueOf method is defined for any Object. In order for it to be called, it will convert all primitive types to their wrapper types and will leave existing objects unchanged. Now it will simply return the value held in the Object reference. So it's pretty simple and it is a way to FORCE AUTOBOXING. You are forcing the conversions I was mentioning earlier.
To answer your second question
Displaying the unfiltered result is simple. Use console.dir().
Look here.
({}).valueOf.call(myvar);
It is the exact equivalent of Object.prototype.valueOf.call(myVar);. Now you already know what valueOf does.
Assuming you know the way Function.prototype.call works, your statement will call the valueOf method in the scope of the object you pass as a this argument to Function.prototype.call(the first parameter is the this object reference).
var myvar = {
"name": "name"
};
({}).valueOf.call(myVar);
// is equivalent to
myVar.valueOf();

explaining this declaration of the global namespace

I see this quite a lot at the top of scripts but I'm not completely sure what it means, can anyone explain?
var whatevername = whatevername || {};
It uses the OR operator to set default values. If whatevername has been set it will be used, otherwise a new empty object will be used.
An example:
function sayHi(options){
options = options || {};
if (options.useAlert){
alert("hi");
} else {
console.log("hi");
}
}
In this case you can always use the options parameter, even if it isn't passed to the function:
sayHi();
sayHi({"useAlert": true});
In the first case {} will be used and options.useAlert will be undefined. In the if statement that's the same as it being set to false and console.log will be used to print.
The OR operator is usually used like this:
if (hasAnEmailAddress || hasAPhoneNumber) {contactPerson()}
If hasAnEmailAddress is true the operator will return the value of hasAnEmailAddress instead of hasAPhoneNumber. If it isn't true the value of the second argument, hasAPhoneNumber will be returned.
That logic is used when setting default values: If the first argument is falsy return the second argument - even if it isn't a boolean value.
it initializes whatevername with an empty object if whatevername hasn't already been initialized.
Equivalent code
if(!whatevername) whatevername = {}
In a lot of languages, you will see this done with a ternary operator, which I think makes it very clear what's going on. Example:
var whatevername = (whatevername != NULL) ? whatevername : {};
In Javascript, if the || operator evaluates to truthy, it will not return a boolean value as one might expect, but the value of the operand which was last evaluated. Therefor, if whatevername is null, it will return a new object, otherwise it will return whatevername. Ruby does this as well, just to name another example of this behaviour.
This is a default value statement. || is the symbol for OR, as you probably know.
The statement reads "set whatevername to whatevername OR an empty object". The OR will pick the first of the two objects that reads to a truthy value (not empty, not false). If whatevername was set, you'll get whatevername. If not (or if set to null), you'll get an empty object.

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