Why Does My Original Array Change Even Though I Create Another One?

If you’re reading this, chances are you’ve stumbled upon a frustrating phenomenon in JavaScript: your original array is changing even though you’ve created a new one. You’re not alone! This is a common gotcha that catches many developers off guard. In this article, we’ll explore the reasons behind this behavior and provide practical solutions to avoid it.

The Problem: Mutating the Original Array

Let’s start with an example. Suppose you have an original array `fruits` and you want to create a new array `moreFruits` that includes all the elements of `fruits` plus some additional ones. You might do something like this:

let fruits = ['apple', 'banana', 'orange'];
let moreFruits = fruits;
moreFruits.push('mango', 'papaya');
console.log(fruits); // Output: ['apple', 'banana', 'orange', 'mango', 'papaya']
console.log(moreFruits); // Output: ['apple', 'banana', 'orange', 'mango', 'papaya']

Wait, what?! You didn’t intend to modify the original `fruits` array, but somehow it got changed. This is because when you assign `moreFruits = fruits`, you’re not creating a new array, but rather a new reference to the same array. Both `fruits` and `moreFruits` are pointing to the same location in memory.

The Solution: Creating a Shallow Copy

To avoid mutating the original array, you need to create a shallow copy of it. A shallow copy creates a new array and copies the references of the original elements to the new array. There are several ways to do this:

• let moreFruits = [...fruits]; (using the spread operator)
• let moreFruits = Array.prototype.slice.call(fruits); (using the slice method)
• let moreFruits = fruits.slice(); (using the slice method directly on the array)

Let’s see how these solutions work:

let fruits = ['apple', 'banana', 'orange'];
let moreFruits = [...fruits];
moreFruits.push('mango', 'papaya');
console.log(fruits); // Output: ['apple', 'banana', 'orange']
console.log(moreFruits); // Output: ['apple', 'banana', 'orange', 'mango', 'papaya']

Ah, much better! Now `fruits` remains unchanged, and `moreFruits` has the additional elements.

The Gotcha: Mutating Objects Within the Array

But wait, there’s more! Even with a shallow copy, you can still mutate the objects within the array. Suppose you have an array of objects:

let people = [
  { name: 'John', age: 30 },
  { name: 'Jane', age: 25 }
];
let morePeople = [...people];
morePeople[0].name = 'Jonas';
console.log(people); // Output: [{ name: 'Jonas', age: 30 }, { name: 'Jane', age: 25 }]
console.log(morePeople); // Output: [{ name: 'Jonas', age: 30 }, { name: 'Jane', age: 25 }]

Ouch! You didn’t intend to change the `name` property of the first object in the original `people` array, but it got modified. This is because the objects within the array are still referencing the same memory location.

The Solution: Creating a Deep Copy

To avoid mutating the objects within the array, you need to create a deep copy of the array. A deep copy creates a new array and recursively copies the objects within the array. One way to do this is using the `JSON.parse(JSON.stringify())` trick:

let people = [
  { name: 'John', age: 30 },
  { name: 'Jane', age: 25 }
];
let morePeople = JSON.parse(JSON.stringify(people));
morePeople[0].name = 'Jonas';
console.log(people); // Output: [{ name: 'John', age: 30 }, { name: 'Jane', age: 25 }]
console.log(morePeople); // Output: [{ name: 'Jonas', age: 30 }, { name: 'Jane', age: 25 }]

Ah, finally! Now `people` remains unchanged, and `morePeople` has the modified object.

Best Practices

To avoid these gotchas, follow these best practices:

1. Always create a shallow copy when you need to modify an array.
2. Use a deep copy when you need to modify objects within an array.
3. Avoid directly modifying an array or object that you didn’t intend to change.
4. Use immutable data structures whenever possible.

Conclusion

In conclusion, the reasons behind your original array changing even though you created another one boil down to JavaScript’s reference-based assignment and the nuances of shallow and deep copying. By understanding these concepts and following best practices, you can avoid these common pitfalls and write more predictable, reliable code.

Solution	Example	Description
Shallow Copy	let moreFruits = [...fruits];	Creates a new array and copies the references of the original elements.
Deep Copy	let morePeople = JSON.parse(JSON.stringify(people));	Creates a new array and recursively copies the objects within the array.

Remember, a little caution and knowledge can go a long way in avoiding these common JavaScript gotchas. Happy coding!

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