Javascript object getter

Question

I tried to execute this small part of code in my browser console:

a = {b: 1}
Object.defineProperty(a, 'b', { get: function() { console.log(5); } })
c = a.b
c
c

Everytime i used to call 'c' varaible, i am not getting console.log(5) to be executed. That means my getter is not calling. I think that might be because i am not calling a.b directly but using c variable instead. Can someone confirm/reject my thoughts ? Also how can i make it call getter everytime i am calling 'c' ?

Thanks.

You have a.b assigned to c. That is when your getter in invoked. And once the get value is assigned to c. c behaves like any other variable. — Nandu Kalidindi
– Nandu Kalidindi, Commented Nov 21, 2017 at 16:22

Suren Srapyan · Accepted Answer · 2017-11-21 16:34:17Z

2

Can someone confirm/reject my thoughts

You will get console.log(5) run every time when you call a.b. But when you use c = a.b, this runs the getter function, copies the result into the c (which is undefined) and starting from here c has its independent value which is returned via a.b. So after this every time when you call c, you evaluate c and get the value of it, not the a.b. In few words c has nothing anymore with a.b.

Also how can i make it call getter everytime i am calling 'c' ?

Actually there is no solution. One thing you can do is returning a function which logs the result, but now you will need to invoke c to get it work. Maybe you can chagne your logic in somewhere, so this is not a good workaround.

const a = {b: 1}
Object.defineProperty(a, 'b', { get: function() { 
      console.log(5); 
      return () => console.log(5);
   } 
});
const c = a.b;
c();
c();

edited Nov 21, 2017 at 16:34

answered Nov 21, 2017 at 16:22

Suren Srapyan

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Comments

TimoStaudinger · Accepted Answer · 2017-11-21 16:22:45Z

0

You're only getting b once, and assigning the returned value, in this case undefined, to c.

The getter is not called anymore if you work with c after that.

answered Nov 21, 2017 at 16:22

TimoStaudinger

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