为传递给函数的数组赋值

Question

var arrN = [1, 2, 3];

function init(arr){
   arr = [];
   console.log(arrN) //output [1, 2, 3], expect []  
}

init(arrN);

When using splice or push methods the array passed to a function is being modified. So I am trying to understand what is happening when using assignment operator, why it's not changing the array? is it creating the local var of the passed array?

Any help will be appreciated!

Answer 1

You need to distinguish between the variable and the actual object (the array). splice and push are changing the object. arr = [] is just changing the variable, the old object just stays as it is.

Answer 2

There is a difference in assigning a different object to a variable or mutating the object currently referenced by a variable.

(Re)assigning

When you do an assignment like:

arr = []; // or any other value

... then the value that arr previously had is not altered . It is just that arr detaches from that previous value and references a new value instead. The original value (if it is an object) lives on, but arr no longer has access to it.

Side note: if no other variable references the previous value any more, the garbage collector will at some point in time free the memory used by it. But this is not your case, since the global variable arrN still references the original value.

Mutating

It is another thing if you don't assign a value to arr , but apply instead a mutation to it, for instance with splice , push , pop , or an assignment to one of its properties, like arr[0] = 1 or arr[1]++ . In those cases, arr keeps referencing the same object, and the changes are made to the object it references, which is visible to any other variable that references the same object, like arrN .

Clearing an array

Now if you want to clear the array that is passed to your function, you must avoid to make an assignment like arr = ... . Instead, use methods that mutate the array in place:

arr.splice(0)

Or, alternatively:

arr.length = 0;

Now you have actually mutated the array, which is also the array referenced by arrN .

Answer 3

In JavaScript, particularly when working with objects the assignment operator ( = ) has three jobs.

1. Assignment
2. Creating a new reference if the right side is an object.
3. Breaking any previous referencing and creating multiple assignments if both sides are objects.

Such as;

var a = [1,2,3],   // a is assigned an array
    b = a;         // b is assigned just a reference to a
a = ["a","b","c"]; // b to a referral is now broken
                   // a is assigned with ["a","b","c"]
                   // b is assigned with [1,2,3]

same would apply if it was done in reverse order;

var a = [1,2,3],   // a is assigned an array
    b = a;         // b is assigned just a reference to a
b = ["a","b","c"]; // b to a referral is now broken
                   // b is assigned with ["a","b","c"]
                   // a keeps it's existing assignment as [1,2,3]

Answer 4

Your are passing arrN to the console instead of passing arr . Also, you just call a function by its name, not by the keyword function . Here is the corrected code:

var arrN = [1, 2, 3];

function init(arr){
   arr = [];
   console.log(arr) 
}

init(arr);

You have also declared init() with an argument arr , which is not needed in this case. What ever the value you pass to init() , the value of arr will be [] because you are reassigning it in the function.