I want to understand this concept which confuses me.
suppose I have the below logic:
function test(){
var jsonObj_1 = {};
var jsonObj_2 = {};
jsonObj_2 = jsonObj_1;
jsonObj_2.myKey = 3;
console.log(jsonObj_2) // result => {myKey:3}
console.log(jsonObj_1) // result => {myKey:3}
}
my question is why jsonObj_1 is equal to {myKey:3} when it's never get assigned?!
you are assigning the ref of jsonObj_1 in jsonObj_2. In simple words the address of the first variable. I would suggest you to read some docs on call by reference and call by value.
When the non-primitive variables are assigned using "=", the reference will also be copied. This may lead to the mutation for all the variables that a particular object is assigned to.
Try the following.
You can use JSON.stringify to convert it to string and then parse into JSON.parse.
jsonObj_2= JSON.parse(JSON.stringify(jsonObj_1));
You can use spread operator.
jsonObj_2 = {...jsonObj_1}
What you will need is Object.assign
const target = { a: 1, b: 2 };
const source = { b: 4, c: 5 };
const returnedTarget = Object.assign(target, source);
console.log(target);
// expected output: Object { a: 1, b: 4, c: 5 }
console.log(returnedTarget);
// expected output: Object { a: 1, b: 4, c: 5 }
Example here: https://googlechrome.github.io/samples/object-assign-es6/
JavaScript Objects are Mutable. They are addressed by reference, not by value.
If jsonObj_1 is an object, the following statement will not create a copy of jsonObj_1:
var jsonObj_2 = jsonObj_1 ; // This will not create a copy of jsonObj_1.The object jsonObj_2 is not a copy of jsonObj_1 . It is jsonObj_1. Both jsonObj_2 and jsonObj_1 are the same object.
Any changes to jsonObj_2 will also change jsonObj_1, because jsonObj_2 and jsonObj_1 are the same object.
