prototype value does not update for every instance in js, why?

Question

if I set a prototype value and make two instances

then update the prototype value in one instance

now other instance prototype value does not update, why?

code is

var A = function() {

}
A.prototype.u = 2;

var a = new A();

var b = new A();

a.u = 4
alert(a.u) // 4
alert(b.u) // 2

it's so unreasonable, it's prototype value not this value. right?

When doing assignment, the prototype chain isn't used, the property is only searched for on the object itself and if not found, a new property is added, masking the one on the prototype. — RobG
– RobG, Commented Jan 24, 2015 at 8:20
@RobG so I cannot update a prototype value in method, like global var? — Roy
– Roy, Commented Jan 25, 2015 at 11:12
Yes you can, but not the way you are trying to do it. You need to assign to A.prototype.u as in Quentin's answer. — RobG
– RobG, Commented Jan 25, 2015 at 11:40
@RobG do you think about to set prototype = {u:[11]}, then change it in instance, just not to type constructor name and prototype — Roy
– Roy, Commented Jan 25, 2015 at 11:49

Quentin · Accepted Answer · 2015-01-24 08:17:34Z

6

You aren't "updating the prototype value". You are writing the new value to the local object and not to the prototype chain. The local property masks the one higher up the chain.

alert(a.u); looks at a, finds a u and alerts it.

alert(b.u); looks at b, doesn't find a u, looks up the prototype chain, finds a u and alerts it.

Compare:

var A = function() {

}
A.prototype.u = 2;

var a = new A();

var b = new A();

a.u = 4;
A.prototype.u = 6;
alert(a.u);
alert(b.u);

answered Jan 24, 2015 at 8:17

Quentin

949k136 gold badges1.3k silver badges1.4k bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

zerkms Over a year ago

alert(b.u) // 2 <-- 6?

Trash Can Over a year ago

6 is right, since b.u is searched first -> not found -> prototype is then searched -> found -> 6. @Quentin's answer says 2 which is not right

Roy Over a year ago

@Quentin if I write a.x = function(){A.prototype.u = 6} it doest not work, why?

Quentin Over a year ago

@Roy — Because you are either still masking it locally or because you don't call the method before you test the value of u.

RobG Over a year ago

@Roy—that just assigns a function to a.x. Until you call it (i.e. a.x()), the value of A.prototype.u won't change.

|

Collectives™ on Stack Overflow

prototype value does not update for every instance in js, why?

1 Answer 1

6 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Your Answer

Sign up or log in

Post as a guest

Related