Does Javascript Slice Method Return A Shallow Copy?

In a Mozilla developer translated Korean lang says 'slice method' returns a new array copied shallowly. so I tested my code. var animals = ['ant', 'bison', 'camel', 'duck', 'elepha

Solution 1:

slice does not alter the original array. It returns a shallow copy of elements from the original array.

Elements of the original array are copied into the returned array as follows:

For object references (and not the actual object), slice copies object references into the new array. Both the original and new array refer to the same object. If a referenced object changes, the changes are visible to both the new and original arrays.

For strings, numbers and booleans (not String, Number and Boolean objects), slice copies the values into the new array. Changes to the string, number or boolean in one array do not affect the other array. If a new element is added to either array, the other array is not affected.(source)

In your case the the array consists of strings which on slice would return new strings copied to the array thus is a shallow copy. In order to avoid this use the object form of array.

Solution 2:

strings are primitive types in JavaScript, so you will get a new array with new strings inside.

Your test array should be an array of objects:

var animals = [{name: 'ant'}, {name: 'bison'}, {name: 'camel'}, {name: 'duck'}, {name: 'elephant'}];

var t = animals.slice(2,4);
console.log(t);

t[0].name = 'aaa';
console.log(t);
console.log(animals);

Solution 3:

The slice method doesn't change the original array or string. It only cuts a portion of the original string or array and returns it as a copy. For more understanding of it, kindly check this video below: https://youtu.be/mUH8hPQfMbg [Slice method made easy for absolute beginners]

Solution 4:

May be you are looking for this. Try this!

let animals = ['ant', 'bison', 'camel', [1, 2]];

let t = animals.slice();

t[0] = 'aaa';    // string (primitive datatype)
t[t.length-1][0] = 0;    // array (object)console.log(t);
console.log(animals);

In case of a shallow copy-

• Objects will reflect change in the original place from where they were shallowly copied because they are stored as references (to their address in the Heap).
• Primitive data types will NOT reflect change in the original place because they are directly stored in the callstack (in Execution Contexts).