Question

I have lived under the assumption that there are primitive types and reference types in Javascript. On a day-to-day basis, I’ve never had this impact me but I was just starting to a lot more JS and wanted to update my ‘thinking’. In other words, I would have betted $20 that the following would return 68

var my_obj = {};
var tmp_obj = {};

tmp_obj.my_int = 38;
my_obj.tmp_val = tmp_obj.my_int;
tmp_obj.my_int = 68;

alert('68 means reference, 38 means primitve: ' + my_obj.tmp_val);

but it returns 38.

Are all instances of numbers primitive types even if they exist in the context of a reference type? If y, I’m really surprised and find that odd behavior(and would be out $20). Or is my example not demonstrating what I think it is?

thx in advance

UPDATE #1

Wow, thx for all the answers. Here’s a slight change which helps me a lot in understaning:

var my_obj={};
var tmp_obj={};
var my_obj_2=tmp_obj;
tmp_obj.my_int=38;
my_obj.tmp_val=tmp_obj.my_int;
tmp_obj.my_int=68
alert('68 means reference, 38 means primitve: ' + my_obj.tmp_val);   // 38
alert('68 means reference, 38 means primitve: ' + my_obj_2.my_int);  // 68
my_obj_2.my_int=78;
alert(tmp_obj.my_int); // tmp_obj is now 78 ie two way

Answer 1

You example would work as expected if you had

     my_obj = tmp_obj;

Then, all the properties would point to the same reference since there would only be one object.

But when you write

     my_obj.tmp_val = tmp_obj.my_int;

then my_obj.tmp_val will take the value that’s stored in tmp_obj.my_int but that doesn’t create a new reference between the 2 objects.