JavaScript - How to modify 'variable' inside function so that it does not revert to origin-CodePudding

I have a script like this:

function changeValue(x){
    x = 30;
}

let a = 3;
changeValue(a);
console.log(a);

The above code outputs 3, but the expected output is 30.
Now I'm aware why this is happening, but how do I modify my code so that the changes made do not revert after the changeValue function ends?
I have read that there are no explicit pointers in JS and I don't want to alter the value using the global variable name or by using the object approach. Is there any other way to solve the problem?
Thank you

CodePudding user response：

The best way would be to use return, because x isn't an object, it's just a variable. It is discarded after the function.

Using Return:

function changeValue(x){
    return 30;
}

let a = changeValue(3);
changeValue(a);
console.log("Output Is:", a);

CodePudding user response：

In JavaScript, all primitive types (integers, floats, etc) are passed by value, so as other people said, for your specific use case, it's better to return the value like this:

function returnValue(x) {
  return x * 10
}

let a = 3;
a = returnValue(a);
console.log(a);

However, if you really want to change the argument and NOT return, you need to do it with pass by reference. As I said above, it's not possible with primitive data types but is possible with objects:

function changeValue(x) {
  x.value = x.value * 10;
}

let a = { value: 3 };
changeValue(a);
console.log(a.value);