What is the reason why JS push objects by reference?

Question

What is the reason why Javascript push objects by reference and not by value?

And why this behavior is only for objects and not also for built-in primitives types?

For example:

let x = {a: 10, b: 100 }
let ar = [];

ar.push(x);
x.a = 9;
console.log(ar[0].a); // This print 9, not 10

I think an answer to this question is useful to understand some deep functions about this language.

Thanks in advance.

Everything in Javascript, not only primitives, is passed by value. — Gerardo Furtado, Dec 30 '19 at 00:50
@GerardoFurtado check this post: [Do objects pushed into an array in javascript deep or shallow copy?](https://stackoverflow.com/questions/8660901/do-objects-pushed-into-an-array-in-javascript-deep-or-shallow-copy/8661088#8661088) — Stefano, Dec 30 '19 at 00:53
I know this, thanks anyway. It's still by value... It's complicated at first, I agree. — Gerardo Furtado, Dec 30 '19 at 00:56
It's very difficult to do otherwise and still maintain some standard sense of object orientation. If values were immutable and functions were free of side-effects, then this would be reasonable -- but then you're much of the way toward functional programming. Javascript makes it easy enough to do OOP, and possible, but less easy to do FP; that was one of the essential design decisions of the language. — Scott Sauyet, Dec 30 '19 at 01:01

score 3 · Accepted Answer · answered Dec 30 '19 at 01:30

3

Everything in JavaScript is passed by value...

But, those values can be a reference value.

The different seems subtle, but its very important.

For example, if you have a function:

function nukeArray(a) {
  a = [];
}

Then, when you call it, a will receive a value that happens to be a reference. That value is immediately discarded and a is assigned a new reference.

If a was a pure reference, then changing a would also change the value of its caller.

In a language like C you can easily pass a reference -- change the reference -- then the caller's value will be changed as well.

#include <stdio.h>
#include <stdlib.h>

void t(char **t) {
    *t = malloc(50);
}

int main() {
    char *tmp;
    printf("pre: %p\n", tmp);
    t(&tmp);
    printf("post: %p\n", tmp);
}

answered Dec 30 '19 at 01:30

Jeremy J Starcher

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Your explanation is clear. But what about my example? If I change `x` object it will change also `ar[0]`. Isn't it like it's a reference? – Stefano Dec 30 '19 at 01:37
2

Yes, it s a reference. But a `reference value` -- meaning, it passes a value and that value is a reference. This is *different* from passing by reference. The real reason why you don't pass an object (or array) by value is that making a copy of that object would get expensive, which would slow down your program. – Jeremy J Starcher Dec 30 '19 at 01:47
1

@Stefano if you don't want a reference to the original object you can always clone it (a shallow copy, probably) and push the cloned object. – Gerardo Furtado Dec 30 '19 at 01:49

What is the reason why JS push objects by reference?

1 Answers1