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

[[email protected]…]

var a = new Array(4);

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

This should return 0, shouldn't it?

[scottgonzalez]

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

[anonymous]

@reporter

var a = [4]; console.log($.inArray(4, a)); 0

var a = [1,4]; console.log($.inArray(4, a)); 1

[Rick Waldron]

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

[dmethvin]

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