Change original variable value by changing value of passed function parameter

Question

Is it possible to change the variable value with function like example below? Whenever I change the passed parameter value the original value doesn't change. Why is this happening?

let b = 3; function t(n) { n = 5; } t(b) console.log(b) // still 3

I know this can be done like this, but I am wondering why the example above wont work.

let b = 3; function t() { return 5; } b = t(); console.log(b)

JavaScript, like many other languages uses pass by value, not pass by reference. So what you want is not possible in that exact way. There are workarounds though. However, in such a simple case, returning the value and letting the caller decide how what to do with it seems the best solution. — Felix Kling
– Felix Kling, Commented Jul 1, 2022 at 7:26

Justinas · Accepted Answer · 2022-07-01 07:21:13Z

2

Passing scalar values to function will pass them by value, not by reference.

One solution would be to pass it as object:

let b = {value: 3}; function t(n) { n.value = 5; } t(b); console.log(b.value);

answered Jul 1, 2022 at 7:21

Justinas

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3 Comments

Felix Kling Over a year ago

Objects are also passed by value but the value is a (different kind of) reference. You wouldn't be able to do n = {value: 5} either.

Justinas Over a year ago

@FelixKling Yes, you are right, it's passed as shallow copy - only inner properties can be changed

Felix Kling Over a year ago

It's not a copy, otherwise making changes to it wouldn't have any effect on the "original" object.

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