jQuery.inArray() returns -1 for value that is in array if it is the only element in array

Reported by:	vic.emond@netingredients.com	Owned by:
Priority:	undecided	Milestone:	None
Component:	unfiled	Version:	1.7.2
Keywords:		Cc:
Blocked by:		Blocking:

var a = new Array(4);

$.inArray(4, a) returns -1.

This should return 0, shouldn't it?

Attachments (0)

Change History (4)

resolution:	→ invalid
status:	new → closed

new Array(4) create [,,,] not [4].

@reporter

var a = [4];

console.log($.inArray(4, a)); 0

var a = [1,4];

console.log($.inArray(4, a)); 1

Yes, here you've used a literal. In your original example, you used the new Array() constructor, which creates arrays like...


var a = new Array(1);

a.length; // 1
a[0]; // undefined


var a = new Array(1, 2, 3);

a.length; // 3
a[0]; // 1

The Array constructor creates an array of N length when passed a single numeric argument

alert(new Array(3).join("ha"));

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