jQuery inArray() returns -1 if array contains single element

Question

jQuery inArray returns -1 if the array contains single element.

var a = Array(1);
console.log($.inArray(1,a));

This returns -1. But if the array contains 2 or more elements it works perfectly.

var a = Array(1,2,3);
console.log($.inArray(1,a));

Returns perfect position.

@sudharsan oh.. :) Then i too wont tell you that i had upvoted this question and answer as well.. — Rajaprabhu Aravindasamy
– Rajaprabhu Aravindasamy, Commented May 5, 2014 at 12:28

Florian Margaine · Accepted Answer · 2014-05-05 12:36:18Z

6

Contrary to what you seem to think, Array(1) doesn't create an array with element 1 but an array of size 1. That's a specific behavior you get when you pass only one argument and it's an integer.

From the MDN :

If the only argument passed to the Array constructor is an integer between 0 and 2^32-1 (inclusive), this returns a new JavaScript array with length set to that number.

You should probably almost never use this Array constructor whose strange behavior leads to many bugs and which is mostly useless. Use this :

var a = [1];

edited May 5, 2014 at 12:36

Florian Margaine

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answered May 5, 2014 at 12:11

Denys Séguret

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6 Comments

Rajaprabhu Aravindasamy Over a year ago

never (or almost never) use the Array constructor. what is the reason for that..?

Denys Séguret Over a year ago

@RajaprabhuAravindasamy 1) its behavior is far too surprising. 2) it's mostly useless

Shashikumar Misal Over a year ago

Then how its returning perfect position with this: var a = Array(1,2,3);

Rajaprabhu Aravindasamy Over a year ago

@ShashikumarMisal he explained that in his post. read carefully with patience.

Denys Séguret Over a year ago

@ShashikumarMisal Because then it's equivalent to [1,2,3]. Read the document I link to.

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