Is a shallow copy created by spreading?

Question

Is a shallow copy created by spreading?

According to the example provided in the documentation,

let first:number[] = [1, 2];
let second:number[] = [3, 4];

let both_plus:number[] = [0, ...first, ...second, 5];
console.log(`both_plus is ${both_plus}`);
first[0] = 20;
console.log(`first is ${first}`);
console.log(`both_plus is ${both_plus}`);
both_plus[1]=30;
console.log(`first is ${first}`);
console.log(`both_plus is ${both_plus}`);

The concept of Spreading results in a deep copy because each array maintains its own set of values, as shown below:

both_plus is 0,1,2,3,4,5
first is 20,2
both_plus is 0,1,2,3,4,5
first is 20,2
both_plus is 0,30,2,3,4,5

The Documentation states that: Spreading creates a shallow copy of first and second. How can we interpret this?

typescript ecmascript-6 deep-copy shallow-copy

Answer 1

Answer №1

When dealing with arrays that consist only of primitive data types, a shallow copy and deep copy are essentially the same. The true distinction arises when the array contains objects or other non-primitive elements.

In JavaScript, values are passed by value. This means that when you create a shallow copy of an array using techniques like spread syntax, each individual value within the original array is duplicated into the new array. Any modifications made to these values will not affect the original array, as they are separate entities.

However, if the array holds references to objects rather than directly storing primitive values, things get more complex. Even though a reference from the original array gets copied to the new one during a shallow copy operation, both references still point to the same underlying object. Therefore, changes made to the objects within the new array will reflect in the original array too.

For instance, consider this example:

const objArray = [{foo: "bar"}];
const shallowCopy = [...objArray];

// Modifying the structure of the array does not impact the original.
// Here, the original array retains its single item while the copy gains a second:
shallowCopy.push({foo: "baz"});
console.log("objArray after push:", objArray);
console.log("shallowCopy after push:", shallowCopy);

// Nonetheless, since both shallowCopy[0] and objArray[0] reference the same object,
// any adjustments made to either will be reflected in the other:
shallowCopy[0].foo = "something else";
console.log("objArray after mutation:", objArray);
console.log("shallowCopy after mutation:", shallowCopy);

Answer 2

When dealing with arrays that consist only of primitive data types, a shallow copy and deep copy are essentially the same. The true distinction arises when the array contains objects or other non-primitive elements.

In JavaScript, values are passed by value. This means that when you create a shallow copy of an array using techniques like spread syntax, each individual value within the original array is duplicated into the new array. Any modifications made to these values will not affect the original array, as they are separate entities.

However, if the array holds references to objects rather than directly storing primitive values, things get more complex. Even though a reference from the original array gets copied to the new one during a shallow copy operation, both references still point to the same underlying object. Therefore, changes made to the objects within the new array will reflect in the original array too.

For instance, consider this example:

const objArray = [{foo: "bar"}];
const shallowCopy = [...objArray];

// Modifying the structure of the array does not impact the original.
// Here, the original array retains its single item while the copy gains a second:
shallowCopy.push({foo: "baz"});
console.log("objArray after push:", objArray);
console.log("shallowCopy after push:", shallowCopy);

// Nonetheless, since both shallowCopy[0] and objArray[0] reference the same object,
// any adjustments made to either will be reflected in the other:
shallowCopy[0].foo = "something else";
console.log("objArray after mutation:", objArray);
console.log("shallowCopy after mutation:", shallowCopy);

Answer 3

Answer №2

When creating a shallow copy, the elements from the arrays first and second are simply added to a new array. On the other hand, in a deep copy, all elements from first and second are first duplicated and then included in the new array.

The main difference lies in whether the individual elements are duplicated into a new object before being appended to the new array.

If you test this concept with primitive values such as numbers, it may not be apparent. However, the distinction becomes evident when dealing with objects.

For instance, consider the following scenario:

let first = [{foo: 'bar'}];
let second = [{fizz: 'buzz'}];
let both = [...first, ...second];

Since spreading generates a shallow copy, the objects in question should pass an equality test:

first[0] === both[0]; // true
second[0] === both[1]; // true

Conversely, if spreading produced a deep copy, the equality test would fail:

first[0] === both[0]; // false
second[0] === both[1]; // false

Answer 4

When creating a shallow copy, the elements from the arrays first and second are simply added to a new array. On the other hand, in a deep copy, all elements from first and second are first duplicated and then included in the new array.

The main difference lies in whether the individual elements are duplicated into a new object before being appended to the new array.

If you test this concept with primitive values such as numbers, it may not be apparent. However, the distinction becomes evident when dealing with objects.

For instance, consider the following scenario:

let first = [{foo: 'bar'}];
let second = [{fizz: 'buzz'}];
let both = [...first, ...second];

Since spreading generates a shallow copy, the objects in question should pass an equality test:

first[0] === both[0]; // true
second[0] === both[1]; // true

Conversely, if spreading produced a deep copy, the equality test would fail:

first[0] === both[0]; // false
second[0] === both[1]; // false

Answer 5

Answer №3

Utilizing the spread operator creates a shallow copy

let arr1 = [{a: 1}, {b: 2}];
let arr2 = arr1;

let arr3 = [...arr1, ...arr2];
console.log(arr3); // [{a: 1}, {b: 2}, {a: 1}, {b: 2}]

arr1[1]['b'] = 10;
console.log(arr3); // [{a: 1}, {b: 10}, {a: 1}, {b: 10}]

Answer 6

Utilizing the spread operator creates a shallow copy

let arr1 = [{a: 1}, {b: 2}];
let arr2 = arr1;

let arr3 = [...arr1, ...arr2];
console.log(arr3); // [{a: 1}, {b: 2}, {a: 1}, {b: 2}]

arr1[1]['b'] = 10;
console.log(arr3); // [{a: 1}, {b: 10}, {a: 1}, {b: 10}]

Is a shallow copy created by spreading?

Answer №1

Answer №2

Answer №3

Similar questions

Show pictures stored in S3

The CSS styles are functioning correctly in index.html, but they are not applying properly in the component.html

What is the best way to transfer my static files to the desired output directory in a TypeScript Express application?

Utilizing Object-Oriented Programming to organize and store data within a class object

Using the RabbitMQ consume method in conjunction with the channel.ack function

What is the best method for choosing visible elements within a scrollable container?

What could be causing the "buffer is not a function" error to occur?

The devastation caused by typing errors in TypeScript

"Although the NextJS client-side data is present, it seems to be struggling to

What is the best way to determine the final letter of a column in a Google Sheet, starting from the first letter and using a set of

Restricting number input value in Vue using TypeScript

Leveraging scanner-js within an Angular2 environment

Is Intellisense within HTML not available in SvelteKit when using TypeScript?

Error message in TypeScript class extension: "TypeError: Object.setPrototypeOf expects an object or null, received undefined."

Tips for creating a seamless horizontal scrolling effect in Angular when hovering (automatically)

Issue with playing audio file using HowlerJS

Learn how to manipulate Lit-Element TypeScript property decorators by extracting values from index.html custom elements

Provide the remaining arguments in a specific callback function in TypeScript while abiding by strict mode regulations

Leveraging the useRoute() function with Vue 3 Composition API and Typescript: [Vue alert]: The injection "Symbol(route location)" was not detected when employing Typescript

How to transfer data between components in Angular 6 using a service