jQuery inArray()'s return values

Question

I was reading jQuery's documentation about the inArray method and I can't really understand the meaning of one sentence. After saying that the method returns -1 if the element is not in the array, and 0 otherwise, it reads:

Because JavaScript treats 0 as loosely equal to false (i.e. 0 == false, but 0 !== false), if we're checking for the presence of value within array, we need to check if it's not equal to (or greater than) -1.

I can't understand what the fact that JavaScript treats 0 as loosely equal to false has to do with the need to check the method's return values. Would't I need to do this even if JavaScript didn't work this way?

Darin Dimitrov · Accepted Answer · 2012-05-07 08:50:14Z

8

It basically means that you need to write the test like this:

if ($.inArray(value, array) > -1) {
    // the element is inside the array
}

instead of:

if ($.inArray(value, array)) {
    // the element is inside the array
}

The reason why you shouldn't use the second construct is because if the value is the first element of the array the inArray function will return 0 (because that's the index of the first element of an array) and as explained in the documentation 0 means false:

if (0) {
    // this will never run
}

answered May 7, 2012 at 8:50

Darin Dimitrov

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Comments

Alnitak · Accepted Answer · 2012-05-07 08:51:04Z

6

They're just covering their backsides over the hideous mis-naming of that function¹.

In other words, they're saying that you mustn't do:

if ($.inArray(a, b) == false) {
   // will fail if the element is in index 0
}

¹ A function named inArray would logically (no pun intended) be expected to return a boolean, not a position.

answered May 7, 2012 at 8:51

Alnitak

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1 Comment

Aquarelle Over a year ago

I agree that it's hideously misnamed. It's clear from the function name that it returns a boolean. And yet it doesn't return a boolean. (Sometimes I want to strangle other developers.) We all need to create our own libraries to include a wrapper function that truly (no pun intended) does return a boolean.

rkw · Accepted Answer · 2012-05-07 09:01:27Z

3

It just means inArray should have really been named indexInArray.

This makes it so that a 0 means it's been found in the first element, and not that it wasn't found.

answered May 7, 2012 at 9:01

rkw

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Comments

zerkms · Accepted Answer · 2012-05-07 08:50:45Z

1

Indexes start with 0, so 0 is a valid value in the case like

var arr = ['foo'];
console.log($.inArray('foo', arr));

So jQuery developers decided to use -1 instead of false to prevent obvious errors in case if developer checks the value presence like

if (!$.inArray(..))

answered May 7, 2012 at 8:50

zerkms

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2 Comments

user1301428 Over a year ago

So instead of using if (!$.inArray(..)) I should use if ($.inArray(..) == -1)?

zerkms Over a year ago

@user1301428: if (!$.inArray(..)) --- is incorrect, because it would match both 0 and false cases (if jquery developers decided to use false). So, yes, compare it to -1 to check if the value doesn't present in the array

DadViegas · Accepted Answer · 2012-05-07 08:53:20Z

1

In javascript 0 is false and 1 is True, or you can use true or false. zero is a position of the array is the first item of the array.

if you compare in the if statment if(0) it is considered has false.

but your answer in the context of the array is that the item is in the zero position.

answered May 7, 2012 at 8:53

DadViegas

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Comments

gdoron · Accepted Answer · 2012-05-07 08:55:07Z

1

They tell you don't check for 0 is it's like false or null etc'
it will be a problem if the value is in index 0 of the array(== false)

Check if the value is greater than -1 or !=-1

Example of the problem:

if ($.inArray(20, [20]))
    // won't run the value is in index - 0

if ($.inArray(20, [20]) > -1)
    // Will run!

DEMO

edited May 7, 2012 at 8:55

answered May 7, 2012 at 8:50

gdoron

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