Create a duplicate of array A in JavaScript with the keys transferred to array B, then make modifications to the elements in

Question

Create a duplicate of array A in JavaScript with the keys transferred to array B, then make modifications to the elements in

Currently, I am delving into JavaScript in regards to array assignment and cloning. A puzzling issue arises when attempting to clone the elements of array A into array B using the spread operator "...". Surprisingly, when modifying the elements in array B, the elements in array A also seem to be affected.

For instance:

A = [{id:1, value:3},{id:2, value:1}];
B = [...A];
B[0].value = 4;
console.log(A[0].value);//the output changes to 4, not 3

I've noticed that this issue does not occur when cloning a regular array:

A = [3, 1];
B = [...A];
B[0] = 4;
console.log(A[0]);//the output is still 3

It is known that one purpose of cloning in JavaScript is to prevent two variables from referencing the same memory location. However, why does altering an element in array B reflect on array A as well? If there are any misconceptions on my part, I would appreciate your guidance in clearing them up. Alternatively, if this problem has been addressed and resolved in past discussions, kindly share the relevant link.

javascript arrays object

Answer 1

Answer №1

To better understand how A works, it's helpful to not view it as directly holding the objects at specific indexes, but rather as holding references to those objects:

const A = [*refA*, *refB*];

In this context, refA and refB point to your {id:1, value:3} and {id:2, value:1} objects respectively.

When you create a new array using [...A], each reference is duplicated in the new array but still points to the same underlying objects in memory:

B = [...A]; // `B` is [*refA*, *refB*]

So if you update B[0].value = 4, you're actually modifying the object referenced by B[0] (i.e., *refA*) and changing its value to 4. Since both A and

B</code share the same object references, changes will be reflected in both arrays.</p>
<p>To prevent this shared reference behavior, you can use <code>.map()

to create a new array with distinct object references:

const B = A.map(obj => ({...obj}));

In this new array creation process, we are generating unique objects by spreading the properties of each object into a fresh object. By doing so, updating B[0].value = 4 won't affect the objects held in A.

Answer 2

To better understand how A works, it's helpful to not view it as directly holding the objects at specific indexes, but rather as holding references to those objects:

const A = [*refA*, *refB*];

In this context, refA and refB point to your {id:1, value:3} and {id:2, value:1} objects respectively.

When you create a new array using [...A], each reference is duplicated in the new array but still points to the same underlying objects in memory:

B = [...A]; // `B` is [*refA*, *refB*]

So if you update B[0].value = 4, you're actually modifying the object referenced by B[0] (i.e., *refA*) and changing its value to 4. Since both A and

B</code share the same object references, changes will be reflected in both arrays.</p>
<p>To prevent this shared reference behavior, you can use <code>.map()

to create a new array with distinct object references:

const B = A.map(obj => ({...obj}));

In this new array creation process, we are generating unique objects by spreading the properties of each object into a fresh object. By doing so, updating B[0].value = 4 won't affect the objects held in A.

Answer 3

Answer №2

When it comes to cloning objects, the Spread Operator and Object.assign() both only perform shallow cloning up to one level. For deeper cloning, utilizing a popular library such as <code>loadash

is recommended, which offers a deepClone() function.

Another common technique used for JSON data, excluding functions, is:

let deepClone = JSON.parse(JSON.stringify(nodesArray));

You can give this method a try as well :)

Answer 4

When it comes to cloning objects, the Spread Operator and Object.assign() both only perform shallow cloning up to one level. For deeper cloning, utilizing a popular library such as <code>loadash

is recommended, which offers a deepClone() function.

Another common technique used for JSON data, excluding functions, is:

let deepClone = JSON.parse(JSON.stringify(nodesArray));

You can give this method a try as well :)

Answer 5

Answer №3

The explanation behind this phenomena is rooted in the fact that the initial example contains an array of objects, all of which are reference variables.

When you create a new instance of the array, you are not actually generating new objects. Rather, you are passing references to the same objects that were utilized in the original array.

Answer 6

The explanation behind this phenomena is rooted in the fact that the initial example contains an array of objects, all of which are reference variables.

When you create a new instance of the array, you are not actually generating new objects. Rather, you are passing references to the same objects that were utilized in the original array.

Answer 7

Answer №4

When utilizing Spread, a new reference is created for the array. However, this method only clones one level deep. Now, array A and array B are pointing to distinct references.

Yet, when it comes to objects (1st case), the elements within the array still reference the same object. This distinction is not present with primitives (2nd case). Thus, altering B[0] impacts the first element of array B, which is a primitive value that gets replaced.

Answer 8

When utilizing Spread, a new reference is created for the array. However, this method only clones one level deep. Now, array A and array B are pointing to distinct references.

Yet, when it comes to objects (1st case), the elements within the array still reference the same object. This distinction is not present with primitives (2nd case). Thus, altering B[0] impacts the first element of array B, which is a primitive value that gets replaced.

Answer 9

Answer №5

Why do the elements of Array A also change when I modify B's elements?

This happens because objects typed values are copied by reference, while primitive typed values are copied by value. This means that in your second example, a deep copy is performed (values were copied), while in the first example a shallow copy is done (references were copied).

This is the concept of shallow copy and deep copy.

Now, let's discuss the spread operator: It performs a deep copy if the data is not nested. For nested data, it deeply copies the first level data and makes shallow copies of the nested data.

Therefore, if you need to deep copy your non-primitive data type values, you can use the structured clone algorithm.

Demo :

const A = [{id:1, value:3},{id:2, value:1}];
const B = structuredClone(A);
B[0].value = 4;
console.log(A[0].value); // 3
console.log(B[0].value); // 4

Answer 10

Why do the elements of Array A also change when I modify B's elements?

This happens because objects typed values are copied by reference, while primitive typed values are copied by value. This means that in your second example, a deep copy is performed (values were copied), while in the first example a shallow copy is done (references were copied).

This is the concept of shallow copy and deep copy.

Now, let's discuss the spread operator: It performs a deep copy if the data is not nested. For nested data, it deeply copies the first level data and makes shallow copies of the nested data.

Therefore, if you need to deep copy your non-primitive data type values, you can use the structured clone algorithm.

Demo :

const A = [{id:1, value:3},{id:2, value:1}];
const B = structuredClone(A);
B[0].value = 4;
console.log(A[0].value); // 3
console.log(B[0].value); // 4

Answer 11

Answer №6

To successfully clone an Array, you must perform a deep copy instead of a shallow copy using the ... operator.

Answer 12

To successfully clone an Array, you must perform a deep copy instead of a shallow copy using the ... operator.

Create a duplicate of array A in JavaScript with the keys transferred to array B, then make modifications to the elements in

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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