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javascriptarrays
Abigale YPC
Abigale YPC

Reputation: 11

Why does pushing to one array affect another?

Consider the following code snippet:

var arr1 = "john".split('');
console.log(arr1); // [ 'j', 'o', 'h', 'n' ]
var arr2 = arr1.reverse();
console.log(arr1); // [ 'n', 'h', 'o', 'j' ]
var arr3 = "jones".split('');
arr2.push(arr3);
console.log(arr1); // [ 'n', 'h', 'o', 'j', [ 'j', 'o', 'n', 'e', 's' ] ]

In the last console.log, why does pushing to arr2 affect arr1? Why does arr1 change when it's not modified, only arr2 is modified?

Upvotes: 0

Views: 374

Answers (3)

rafascar
rafascar

Reputation: 343

There are two modifications to arr1 taking place on your code.

var arr2 = arr1.reverse();

The function Array.prototype.reverse() reverses an array in place. This means that arr1 will be reversed itself.

Also, when assigning objects, you are only passing a reference to them, so arr2 is actually arr1 (reversed) after this line.

arr2.push(arr3);

Because you assigned arr2 as a reference to arr1, in this line you are actually modifying arr1, being equivalent to writing arr1.push(arr3).

To create a new object from an array, you can use the function Array.prototype.slice(), since this method returns a copy of an array.

var arr1 = "john".split('');  
console.log(arr1); // [ 'j', 'o', 'h', 'n' ]
var arr2 = arr1.slice().reverse(); 
console.log(arr1); // [ 'j', 'o', 'h', 'n' ]
var arr3 = "jones".split(''); 
arr2.push(arr3);
console.log(arr1);  // [ 'j', 'o', 'h', 'n' ]

Upvotes: 1

Timothy Miller
Timothy Miller

Reputation: 2429

The reverse() method in JavaScript mutates the array. This means that it changes the data in place—in the arr1 variable—and then returns a reference to that array. So arr2 is just a reference to arr1, rather than a new set of data.

If you want completely new data, you can do as they say in this StackOverflow question:

var arr2 = arr1.slice().reverse();

That will slice a new set of data and then reverse it, hence returning a new array instead of a reference to the first array.

Example:

var array = ['a', 'b', 'c', 'd', 'e'];

var notmutated = array.slice();
console.log('not mutated:', notmutated);

var mutated = array.reverse();
console.log('mutated: ', array);

Upvotes: 2

Anderson Marques
Anderson Marques

Reputation: 898

var arr1 = "john".split('');  
console.log(arr1); // [ 'j', 'o', 'h', 'n' ]
var arr2 = arr1.slice().reverse(); 
console.log(arr2); // [ 'n', 'h', 'o', 'j' ]
var arr3 = "jones".split(''); 
arr2.push(arr3);
console.log(arr1);  // [ 'j', 'o', 'h', 'n' ]
console.log(arr2);  // [ 'n', 'h', 'o', 'j', ['j','o','n','e','s'] ]
  1. split() function return a new array.
  2. reverse() function does not return a new array. So, arr1 is arr2.
  3. push() function does not return a new array. So, arr1 and arr2 have added the arr3 as the last element.

Use slice() function to make a copy of your array.

References:

split function

reverse function

push function

Upvotes: -1

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