Javascript prototype shared values confusion

Question

I have the following code:

function Bar() {
}

Bar.prototype.c = 0;

var foo = new Bar();
foo.c = 20;

var test = new Bar();
console.log(test.c);

The value I get in the console is 0. I thought that these objects would share the same prototype object, thus the output would be 20. Why is this not so?

Its called shadowing, providing a value on the instance or closer to the instance in the prototype chain. That and more is explained here: stackoverflow.com/questions/16063394/… mutate a prototype value on an instance and it'll change the prototype. — HMR
– HMR, Commented Nov 30, 2014 at 13:31

SLaks · Accepted Answer · 2014-11-30 13:23:58Z

1

They do share the same prototype.

However, when you set foo.c, you're setting a property on foo, not its prototype.

answered Nov 30, 2014 at 13:23

SLaks

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