The Array Still Has Three Elements
In Javascript, delete looks like a natural way to remove something from an array.
You have an element at index 1. You delete it. Surely the array becomes shorter.
Not quite.
delete removes the property at that index. The array itself stays the same length, leaving a hole behind.
🔴 The Problem
Consider this:
const users = ["Alice", "Bob", "Charlie"];
delete users[1];
console.log(users);
console.log(users.length);
You might expect:
["Alice", "Charlie"]
2
Instead, the array still has a length of 3:
["Alice", empty, "Charlie"]
3
The important part is that index 1 is now missing.
You can prove it:
console.log(1 in users); // false
console.log(users.length); // 3
The array did not shift "Charlie" from index 2 to index 1.
You removed the property at index 1 and left a hole where it used to be.
🟢 The Solution
If you mean "remove this element and shift everything after it," use splice():
const users = ["Alice", "Bob", "Charlie"];
users.splice(1, 1);
console.log(users);
console.log(users.length);
Now the result is what you probably expected:
["Alice", "Charlie"]
2
splice() actually changes the array by removing the element and moving the later elements down.
If you do not want to mutate the original array, modern JavaScript also gives you toSpliced():
const users = ["Alice", "Bob", "Charlie"];
const remaining = users.toSpliced(1, 1);
Now users stays unchanged, while remaining contains the shorter array.
⚡ The Part That Bites
The really interesting part is that a hole is not the same as undefined.
Compare these:
const a = ["Alice", undefined, "Charlie"];
const b = ["Alice", "Bob", "Charlie"];
delete b[1];
Reading index 1 gives undefined in both cases:
console.log(a[1]); // undefined
console.log(b[1]); // undefined
But the index exists in a and does not exist in b:
console.log(1 in a); // true
console.log(1 in b); // false
You can see the same difference with Object.keys():
console.log(Object.keys(a)); // ["0", "1", "2"]
console.log(Object.keys(b)); // ["0", "2"]
That difference can also show up when you iterate.
For example, forEach() skips the hole entirely:
b.forEach((value) => {
console.log(value);
});
This prints only:
Alice
Charlie
There is no callback for the missing index.
That is where a seemingly harmless delete can become a debugging headache. The array can still have the same length, but code iterating over it may no longer see every index you think is there.
🧠 The Takeaway
Arrays are objects with numeric properties and a length property.
So when you write:
delete users[1];
you are deleting a property. You are not asking JavaScript to remove an element and close the gap.
Think of it like this:
delete -> remove the property at that index
splice -> remove the element and shift the array
And if you need an immutable version:
toSpliced -> create a new array without the element
The dangerous part is not that delete fails.
It does exactly what JavaScript asked it to do.
The problem is that what JavaScript means by "delete this array element" is probably not what you meant.
Top comments (1)
The properties are actually still strings, just like other objects. When you access them with a number, type coercion occurs.