Why Does Listing The Actual Constructor Of A Class In Javascript Important
Solution 1:
Every function has a prototype
property, it's assigned when the function object is created, it points to a newly created object that inherits from Object.prototype
, and it has a constructor
property, that simply points back to the function itself.
The purpose of the prototype
property is to give a way to implement inheritance, using constructor functions. When you call a function with the new
operator, it will create a new object that inherits from that constructor's prototype
.
Now, the purpose of the constructor
property is to have a way to refer back to the constructor that has created an object, for example:
function Foo () {}
// default value of the property:
Foo.prototype.constructor == Foo; // true
This property is inherited by the "instances" of Foo
, so you can know which constructor was used to create an object:
var foo = new Foo();
foo.constructor == Foo;
If you assign a new object to the function's prototype, this relationship is lost:
functionBar () {}
Bar.prototype = { inherited: 1 };
Bar.prototype.constructor == Bar; // falseBar.prototype.constructor == Object; // true
And it affects also the instances of the function:
var bar = newBar();
bar.constructor == Bar; // false
bar.constructor == Object; // true
Another similar case is when you have two or more levels of inheritance using constructors, the most common way is used to denote the inheritance relationship between the functions, is to assign the prototype
property of the second level, e.g.:
function Parent() {}
function Child () {}
Child.prototype = new Parent();
The above code has several problems, first, it executes the logic of the parent constructor to create the inheritance relationship, but that's another story, in the above example the constructor
property is also affected, since we replace completely the Child.prototype
object:
var child = new Child();
child.constructor == Parent; // true
If we replace replace the value of the constructor
property of Child.prototype
after assigning it, it will show the expected behavior:
functionChild() {}
Child.prototype = newParent();
Child.prototype.constructor = Child;
var child = newChild();
child.constructor == Child; // true
Solution 2:
I believe this has to do with instantiating Bar with the new keyword. I believe using new will look for Bar.prototype.constructor. Before that line the object linked to Bar.prototype.contructor is of type Foo so when instantiated without that line it will make a Foo object instead of a Bar object.
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