I'm trying to understand how prototypes work. Why does the following break?
var A = function A(){this.a = 0},
aa = new A;
A.prototype = {hi:"hello"};
aa.constructor.prototype //->{hi:"hello"} ok so far :)
aa.hi //undefined?? why? :(
I'm trying to understand how prototypes work. Why does the following break?
var A = function A(){this.a = 0},
aa = new A;
A.prototype = {hi:"hello"};
aa.constructor.prototype //->{hi:"hello"} ok so far :)
aa.hi //undefined?? why? :(
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edited May 21, 2015 at 3:54
hello_there_andy
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asked Dec 30, 2010 at 18:35
JohnJohn
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- 1 woops, made a correction to aa.hi – John Commented Dec 30, 2010 at 18:48
- Removed salutation: "Thanks in advance!", don't do it next time – hello_there_andy Commented May 21, 2015 at 2:38
1 Answer
Reset to default 12I think you meant in the last line aa.hi
instead of aa.hello
.
It gives you undefined
because the A.prototype
is assigned after the new object (aa
) has been already created.
In your second line:
//...
aa = new A;
//...
This will create an object that inherits from A.prototype
, at this moment, A.prototype
is a simple empty object, that inherits from Object.prototype
.
This object will remain referenced by the internal [[Prototype]]
property of the aa
object instance.
Changing A.prototype
after this, will not change the direct inheritance relationship between aa
and that object.
In fact, there is no standard way to change the [[Prototype]]
internal property, some implementations give you access through a non-standard property called __proto__
.
To get the expected results, try:
var A = function A () { this.a = 0 };
A.prototype = { hi:"hello" };
var aa = new A;
aa.hi; // "hello"