With javascript is it possible to console.log or list the .prototype properties associated with an object of a given type?
I'm using a firebase cloud-firestore and would like to see what exists on the prototype for calls to the database.
For example a db call returns an Object of type QuerySnapshot { ... }. I'd like to see what exists on QuerySnapshot.prototype.
I can map through the returned values using .map, but if I console log out the object, it's not an array, so presumably it is inheriting a custom .map from prototype?
Yes, you can use Object.getPrototypeOf to get an object's prototype, Object.getOwnPropertyNames to get its properties that have string names, and Object.getOwnPropertySymbols to get its properties that have Symbol names. You might want to repeat the process (going to the prototype's prototype and so on) to get the properties at every level.
The advantage to getOwnPropertyXYZ is that it includes properties that are non-enumerable, whereas for-in and Object.keys only return enumerable properties. (They also don't include Symbol-named properties.)
Alternately, you can console.log the object and explore its properties with the browser's interactive object explorer in the devtools.
I'm a bit confused by the Array.length property (i.e. a property named length on the Array function object) and array_instance.length (i.e. a property named length on instance of array object)
So what is difference between the two length property and when should/shouldn't we use them?
Edit 1:
there is also a length property on Array.prototype object. I am so confused.
Edit 2
Just to paint a clearer picture, here are the different length properties I have found
Edit 3
This is one of the follow up questions I asked in the comment section, but I think the question is important to fully understanding length property, so I have copy-pasted here in the main section
Follow up question:
Array.prototype.hasOwnProperty('length') and Array_instance.hasOwnProperty('length') return true, does that mean there are two length properties, one on array_instance, one on Array.prototype object, with the array_instance.length overshadowning the Array.prototype.length?
functions have a .length property which corresponds to how many arguments they are expecting. For example:
const unary = (a) => {
}
const binary = (a, b) => {
}
console.log(unary.length);
console.log(binary.length);
So the Array constructor has a length of 1 because it expects one parameter to be passed to it (namely, the size of the array).
array objects also have a .lengthproperty, which is unrelated other than having the same name. This property says how large the array currently is.
I really think a lot of the other answers have covered everything needed here, but as it seems the OP hasn't had what they see a a clear answer, I will try to set everything out, somewhat extensively - but as clearly as I can - in order to clarify. (Apologies if anyone thinks I am "stealing" their answer - I assure you that this is not the intention, and I'm deliberately not looking at them as I type this, but I've certainly read most of them and even upvoted a few.)
Array.length
This has already been well-covered above. Array is a native JS function, which you can use for creating arrays. It's less common then simply defining an array literal (and as far as I know there is no reason it would ever be preferable), but instead of var a = [1,2,3] you are allowed to do this:
var a = Array(1,2,3);
console.log(a);
Note in passing that you don't want to do this to create a singleton array, there is an utterly mad gotcha of a special case when you supply exactly one parameter which happens to be an integer:
var a = Array(5);
console.log(a);
Although that shows what appears to be an array of 5 undefined values in whatever JS console implementation SO uses, that's not actually quite what has been created (nor what is displayed in the current version of Chrome). I'm getting way off-topic but I'll refer you to Kyle Simpson's excellent walkthrough of the madness.
Anyway, since Array is a function, as others have already observed, it has a length property as all functions do. I'm really not sure why it evaluates to 1 though - for a user-defined function it is the number of arguments the function was formally declared with, but since Array like all native JS functions isn't actually implemented in JS, I couldn't tell you how the implementation actually works. Clearly it can take any number of arguments and thus is a rather "exotic" function. I don't think that Array.length would ever be useful, no matter what value was arbitrarily assigned to it, but it seems that most implementations go for 1, even if the specification leaves it open. (I'm not enough of a spec devotee to know if this is actually defined in there or left up to implementations.)
arrayinstance.length
This is just the feature of arrays that we know and use all the time. All JS arrays get this property - note that, although it is a property rather than a method (that is, it is not a function), it nevertheless "auto-updates" as the array gets longer/shorter:
var a = [1,2,3];
console.log(a.length);
a.push(4);
console.log(a.length);
a.pop();
a.pop();
console.log(a);
console.log(a.length);
Although as I said, Javascript's native constructors are not implemented in terms of JS itself, you could implement this kind of behaviour by defining a getter (at least since ES5).
Array.prototype.length
To fully explain what Array.prototype (and similar objects) is would take me deep into how Javascript's object system work. Suffice to say here that, unlike class-based languages (and JS does not have classes, even in ES6, despite the class syntax allowing us often to pretend it does), JS does not have the usual concept of "inheritance" that so-called OO languages do. In JS's version (sometimes called "prototypal inheritance"), what happens it that each object has an "internal prototype" which references some other object - and if you try to access a property on that object which it doesn't have, the JS engine will look at that "prototype object" for the property and use its value instead.
It's actually a very simple mechanism, but there are a number of things in the language which confuse this, one of them being the fact that functions (which are also objects) have a property called prototype - which does not point to the real "prototype" object which gets consulted if a nonexistent property is referenced on the function object. (A normal function foo has Function.prototype is the object that it delegates to - not foo.prototype.) However, if you declare a function foo, an object called foo.prototype is created - which is basically an empty, nondescript object. Its significance is that if the function foo is used as a "constructor" - that is, if you make an object by calling new foo() - foo.prototype will then be the object that JS will look up properties (including methods) on if any object constructed from foo happens to fail a property lookup.
This is why, at least in pre-ES6 code, you quite frequently saw this kind of pattern:
function foo(a,b) {
this.a = a;
this.b = b;
}
foo.prototype.add = function() {
this.a = this.a + this.b;
}
var myFoo = new foo(1,2);
console.log(myFoo.a);
myFoo.add();
console.log(myFoo.a);
myFoo.add();
console.log(myFoo.a);
console.log(myFoo.hasOwnProperty("add"));
Despite appearances, myFoo doesn't actually have a method add in this example - as confirmed by the final console.log. When the JS engine fails to find the property though, it goes to myFoo's "internal prototype", which happens to be foo.prototype. And that's why the method works, as it would on any object constructed from foo.
Anyway, this is leading up to the fact that Arrays, which could be (although almost never are) constructed by calling new Array (I didn't use the new operator above, but I could have done, this is a case where it makes no difference), therefore delegate to Array.prototype. All those array methods that you know and love don't "really" exist on the arrays you call them on:
var a = [1,2,3];
a.push(4);
console.log(a);
console.log(a.hasOwnProperty("push"));
console.log(Array.prototype.hasOwnProperty("push"));
So array methods only work because those methods are actually found on the Array.prototype object, to which all arrays delegate for property/method access if the lookup doesn't succeed on the array itself. (And this is why, if you look up any of them on MDN, the top of the page always says Array.prototype.<method_name>, because that's where the method "really" lives.)
A drastic demonstration of this (please DON'T do this in production code!)
// you're used to doing this, and it works:
[1,2].push(3);
console.log("that went fine");
// vandalism on a grand scale!
Array.prototype.push = undefined;
// now you can'tse the "push" method anymore!
[1,2,3].push(4);
But I'm going to end on something of an anticlimax. The above is true for array methods - but the length array property isn't a method. As observed above, it's just a "plain" (non-function) property, which "magically" behaves somewhat like a function call. As observed in the OP, .length property accesses don't delegate as the method calls shown above do, the property exists on each array in itself.
So why does Array.prototype still itself have a length property? Well, Array.prototype is actually itself an array. In fact, that's not the only thing:
Array.prototype.push(1);
console.log(Array.prototype);
Function.prototype();
notice that Array.prototype.push(1) ends up with Array.prototype being the singleton array [1]. So Array.prototype is "kind of like" the empty array (it's not exactly the same, because it has all those methods mentioned above directly accessible on the object itself, which a "normal" empty array doesn't). And with Function.prototype, although calling it didn't output anything, the fact that no TypeError was raised proves that it really is a function (it's actually a "no-op" function, like function() {}, but once again with various methods diretcly on it - the methods which every function has access to, such as .call and .bind).
Anyway, to cut the digression short, since Array.prototype is - as far as ordinary array properties are concerned, at least - an empty array, this explains why it has a length property, and why it's equal to 0.
I hope this clears things up, as well as demonstrating some of the more intriguing parts of Javascript.
The first part has already been answered, Array constructor is a function and functions have a .length property.
For the second, Array.prototype.length, it's a bit more obscur...
Array.prototype is actually an Array:
console.log(Array.isArray(Array.prototype)); // true
Array.prototype.push('hello');
console.log(Array.prototype.length); // 1
console.log(Array.prototype[0]); // "hello"
As to why is it an Array? Because specs say so:
The Array prototype object is an Array exotic objects and has the internal methods specified for such objects. It has a length property whose initial value is 0 and whose attributes are { [[Writable]]: true, [[Enumerable]]: false, [[Configurable]]: false }..
There is also a note specifying that it is for compatibility reasons with previous versions of the specs.
As to why is was designed as being an Array? I have no really strong idea...
Disclaimer:
I have tried to show how constructor and instance works in general. But in fact, they have huge difference between different constructors.
Any constructor has been set its length with the value specified by the spec. Specially most of them are set to 1:
Number.length // 1
Object.length // 1
Function.length // 1
Array.length // 1
// etc.
Similarly,
Array.constructor.length // 1
// etc.
Like #Kaiido pointed out in the comment, you can see some constructor length is set:
Document.length // 0
Int8Array.length // 3
Date.length // 7
And you may also notice the length of the instance is undefined as they are not set. - It's out of the scope though.
let d = new Date
d.length === undefined // true
See at the bottom for relevant references.
But when you have an instance of it, then you're creating a new value of it:
typeof new Array === typeof Array.prototype // true
typeof new Function === typeof Function.prototype // true
// etc.
So, when you use an instance it has no length value set because it has no any parameters:
let arr = new Array // Array.prototype
arr.length === 0 // true
But when you use an instance with parameter, then you have the length property with the value of parameters
let arr = new Array(20)
arr.length === 20 // true
let func = function(a1,a2,a3,a4){}
func.length === 4 // true
// etc.
So now, you have been wondering why the constructor has length value equal to 1?
It's because the spec has set the value to be 1 initially.
Every built-in Function object, including constructors, has a length property whose value is an integer. Unless otherwise specified, this value is equal to the largest number of named arguments shown in the subclause headings for the function description, including optional parameters.
The value of the [[Prototype]] internal slot of the Object constructor is the intrinsic object %FunctionPrototype%.
Besides the length property (whose value is 1),
See these references:
Standard built in objects,
19.1.2 Properties of the Object Constructor,
19.2.2 Properties of the Function Constructor,
etc.
Also, you can see the prototype has length to 0, you already know it why in the preceding example.
Though, here's just a reference stating that:
19.2.3 Properties of the Function Prototype Object
And there some constructor whose length is set different. This is the out of scope of this answer. Though, here's a reference for the date constructor:
20.3.3 Properties of the Date Constructor
So, it's totally up to the spec how they have been defined.
Array.length
Array is constructor which means its type is "function". You try the checking it console.
typeof Array //"function"
According to MDN
The length property indicates the number of parameters expected by the function.
As the Array function expects single argument so Array.length = 0
array_instance.length
The length property of an object which is an instance of type Array sets or returns the number of elements in that array
As we know that arrays are actually objects so objects can have properties. The property length is on the instance of array.
Now second question you may ask why we don't get the length properties of array using Object.keys or for..in loop. The answer is because this property is not Enumerable.
let arr= [];
//this.property 'length2' will not be shown in for..in or Object.keys();
Object.defineProperty(arr,'length2',{
value:'length2 for test',
enumerable:false
})
//this property 'x' will be displayed in for..in loop and Object.keys()
Object.defineProperty(arr,'x',{
value:'length2 for test',
enumerable:true
})
console.log(Object.keys(arr)); //["x"]
Array.prototpe.length
According to the DOCS
The initial value of Array.prototype.constructor is the standard built-in Array
The Array prototype object is itself an array; its [[Class]] is "Array", and it has a length property (whose initial value is +0) constructor
Actually Array.prototype is an array. And remember array is always object. So it can have properties. The methods of the Array are stored in form of key:value. and there is no element in that array so it Array.prototype.length returns 0. If you push() some elements into it you will see it as array.
console.log(Array.prototype.length) //0
console.log(Array.isArray(Array.prototype)) //true
//adding element to array
Array.prototype.push('x')
console.log(Array.prototype.length) //1
As I explained in second parts you can hide properties of Object by setting enumerable:false. All the methods are keys of Array.prototype But now shown in for..in loops.
Array.length
For the number of properties in the array or the length property of an object which is an instance of type Array sets or returns the number of elements in that array.
Array.prototype.length
Inherited number of properties in the array. When you check Array.length you're actually checking Array.prototype.length
So I was playing abound in the V8 console and I did
Object.getOwnPropertyNames([])
I expected to get [] as a result, however ["length"]
SO this means that instead of being part of the prototype chain, length is a member property of all Array objects.
Is this a bug, or is there any design or specific reason length is not the part of a prototype chain?
Prototype properties are shared across objects. So if length is put on prototype, all the array objects will have same length, which is wrong. Length signifies number of elements in current array and should remain property of self.
That is not a bug. By definition, Object.getOwnPropertyNames will return all the enumerable and non-enumerable own properties of an object. When it comes to an array, length is an own property but its enumerable property is false. That's the reason why it is getting included in the result.
you can test it with the following snippet,
console.log(Object.getOwnPropertyDescriptors([]));
The above code will return the descriptors of all the own properties. Inspect it, you will get to know about its enumerable property.
Because it is not enumarable (Object#propertyIsEnumerable).
Further reading: Enumerability and ownership of properties
console.log([].propertyIsEnumerable('length'));
I need to delete the property of Object. Given an "id", I must delete value[id]. I try this code:
delete value[id];
But the delete operator deletes only a reference, never an object itself.Anyone can suggest me any methods to delete forever a objects property?
JavaScript doesn't allow you to do such a thing, it is garbage collected, this means you have no direct control over what happens in memory. You can only delete a reference. Make sure you delete it anywhere else it is used if you want it gone forever.
FROM Mozilla
go through the link, it explains it properly..
Main points from the link is added below..
Unlike what common belief suggests, the delete operator has nothing to do with directly freeing memory (it only does indirectly via breaking references. See the memory management page for more details).
If the delete operator succeeds, it removes the property from the object entirely. However, if a property with the same name exists on the object's prototype chain, the object will inherit that property from the prototype.
delete is only effective on an object's properties. It has no effect on variable or function names.
While sometimes mis-characterized as global variables, assignments that don't specify an object (e.g. x = 5) are actually property assignments on the global object.
delete can't remove certain properties of predefined objects (like Object, Array, Math etc). These are described in ECMAScript 5 and later as non-configurable.
I am actually not sure if I just stumbled upon an unwanted behavior in javascript or if this is somehow intended behavior.
The following code results in a true statement:
var test= {"test":1}
document.write("constructor" in test);
http://jsfiddle.net/xyatxm2g/2/
If I change it to the following code, it returns false as it should:
var test= {"test":1}
document.write(test.hasOwnProperty("constructor"));
http://jsfiddle.net/fg06ovvc/2/
The hasOwnProperty method, as the name says, look into the object to see if it has the property itself.
But when you use 'propertyName' in test, you're not only looking into the object's own properties, but also the properties that come from inheritance.
In that case, constructor is a property that resides inside the Object's prototype, so all objects have that property, because they all inherit from Object.
Quote from MDN
Every object descended from Object inherits the hasOwnProperty method.
This method can be used to determine whether an object has the
specified property as a direct property of that object; unlike the in
operator, this method does not check down the object's prototype
chain.
From the MDN documentation:
Inherited properties
The in operator returns true for properties in the prototype chain.
"toString" in {}; // returns true
Whereas the hasOwnProperty() method only checks for properties directly on the object, not inherited (i.e. not on the prototype chain).
Following MDN documentation, it is not an enumerable field.
https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Object/propertyIsEnumerable
You can perform following test:
var obj = {"t": 23};
obj.propertyIsEnumerable("t")
results: true
obj.propertyIsEnumerable("constructor")
results: false
There is a complete example on this document, under section:
Direct versus inherited properties
I think it may be normal behaviour here that the key in object operator is searching through the prototype chain and returning true for Object.prototype.constructor. See this discussion - it goes over a related topic.
How do I check if an object has a property in JavaScript?