Includes() vs indexOf() in JavaScript

#javascript #arrayprototype

ES2016 Specifications included theÂ includes() methodÂ forÂ Array data structure.Â The includes() method check if an array includes a certain element, returning true or false as appropriate.
But in ES5 we are used to performing operations like this with indexOf() method.

Using includes() method.

const array = [1,2,3,4,5,6];

if(array.includes(4) ){
console.log("true 4 was found in the array")// true 4 was found in the array
}

Let's perform the same operation with indexOf() method.

const array = [1,2,3,4,5,6];

if(array.indexOf(4) > -1 ){
console.log("true 4 was found in the array")// true 4 was found in the array
}

Using includes() method to check for NaN

Â constÂ  array = [NaN];

if (array.includes(NaN)){
console.log("true. NAN was found in the array");// true. NAN was found in the array
}

This is where things begin to fall apart with indexOf() method.

const  array = [NaN];
if (array.indexOf(NaN) == -1){
    console.log("NaN not found in the array");//NaN not found in the array
}

Checking for undefined with the includes() method.

const array = [, , , ,];

if(array.includes(undefined)){
console.log("true array elements are undefined");// true array elements are undefined
}

Let's see how indexOf() method will handle this operation.

const array = [, , , ,];

if(!array.indexOf(undefined) == -1 ){
console.log("true. array elements are undefined");
}else {
console.log("Sorry can't find undefined");// Sorry can't find undefined
}

The includes() methodÂ does not distinguish between -0 and +0

const a = [-0].includes(+0);
console.log(a);//true

Typed Arrays will also have a method includes()

let array = Uint8Array.of(2,6,4);
console.log(array.includes(4));//true

Summary

The includes method finds NaN and undefined whereas the indexOf method doesn't.
The includes() methodÂ does not distinguish between -0 and +0(This is not a bug, but clearly how javascript works. Check javascript Number type)
Read more from MDN about Array.prototype.includes()

Top comments (9)

Jonas Winzen • Sep 5 '17

Nice article.
I have one correction for you.

The line:
if(array.indexOf(4) >= -1 )

Needs to be:
if(array.indexOf(4) > -1 )

In order to check, whether an element is in the array. Otherwise the condition would always be true.

John Samuel Obinna • Sep 5 '17

Yeah. indexOf() always return -1 if element is not found. Thanks Jonas

redeemefy • Nov 29 '19 • Edited

I'm wondering about performance. Meaning if includes or indexOf keep searching the array even though already found the element. MDN docs doesn't say if includes breaks the search once finds an element.
developer.mozilla.org/en-US/docs/W...