How do I empty an array in JavaScript?

Answer 1

Array.prototype.clear = function() {
    this.length = 0;
};

And call it: array.clear();

Answer 2

A.splice(0);

I just did this on some code I am working on. It cleared the array.

Answer 3

I'm surprised no one has suggested this yet:

let xs = [1,2,3,4];
for (let i in xs)
    delete xs[i];

This yields an array in quite a different state from the other solutions. In a sense, the array has been 'emptied':

xs
=> Array [ <4 empty slots> ]

[...xs]
=> Array [ undefined, undefined, undefined, undefined ]

xs.length
=> 4

xs[0]
=> ReferenceError: reference to undefined property xs[0]

You can produce an equivalent array with [,,,,] or Array(4)

Answer 4

Ways to clear an existing array A:

Method 1

(this was my original answer to the question)

A = [];

This code will set the variable A to a new empty array. This is perfect if you don't have references to the original array A anywhere else because this actually creates a brand new (empty) array. You should be careful with this method because if you have referenced this array from another variable or property, the original array will remain unchanged. Only use this if you only reference the array by its original variable A.

This is also the fastest solution.

This code sample shows the issue you can encounter when using this method:

var arr1 = ['a','b','c','d','e','f'];
var arr2 = arr1;  // Reference arr1 by another variable 
arr1 = [];
console.log(arr2); // Output ['a','b','c','d','e','f']

Method 2 (as suggested by Matthew Crumley)

A.length = 0

This will clear the existing array by setting its length to 0. Some have argued that this may not work in all implementations of JavaScript, but it turns out that this is not the case. It also works when using "strict mode" in ECMAScript 5 because the length property of an array is a read/write property.

Method 3 (as suggested by Anthony)

A.splice(0,A.length)

Using .splice() will work perfectly, but since the .splice() function will return an array with all the removed items, it will actually return a copy of the original array. Benchmarks suggest that this has no effect on performance whatsoever.

Method 4 (as suggested by tanguy_k)

while(A.length > 0) {
    A.pop();
}

This solution is not very succinct, and it is also the slowest solution, contrary to earlier benchmarks referenced in the original answer.

Performance

Of all the methods of clearing an existing array, methods 2 and 3 are very similar in performance and are a lot faster than method 4. See this benchmark.

As pointed out by Diadistis in their answer below, the original benchmarks that were used to determine the performance of the four methods described above were flawed. The original benchmark reused the cleared array so the second iteration was clearing an array that was already empty.

The following benchmark fixes this flaw: http://jsben.ch/#/hyj65. It clearly shows that methods #2 (length property) and #3 (splice) are the fastest (not counting method #1 which doesn't change the original array).

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This has been a hot topic and the cause of a lot of controversy. There are actually many correct answers and because this answer has been marked as the accepted answer for a very long time, I will include all of the methods here. If you vote for this answer, please upvote the other answers that I have referenced as well.

Answer 5

Use below if you need to empty Angular 2+ FormArray.

public emptyFormArray(formArray:FormArray) {
    for (let i = formArray.controls.length - 1; i >= 0; i--) {
        formArray.removeAt(i);
    }
}

Answer 6

If you are using

a = []; 

Then you are assigning new array reference to a, if reference in a is already assigned to any other variable, then it will not empty that array too and hence garbage collector will not collect that memory.

For ex.

var a=[1,2,3];
var b=a;
a=[];
console.log(b);// It will print [1,2,3];

or

a.length = 0;

When we specify a.length, we are just resetting boundaries of the array and memory for rest array elements will be connected by garbage collector.

Instead of these two solutions are better.

a.splice(0,a.length)

and

while(a.length > 0) {
    a.pop();
}

As per previous answer by kenshou.html, second method is faster.