javascript doesn't correctly return from the loop [duplicate]

Question

Possible Duplicate:
jQuery each method does not return value

It seems the return statement does not break the loop and return the function. The following code keeps return false....

var __in__ = function(elem, array){
    // $.each(array, function(index, item) {  
   array.forEach(function(index, item) {
        if (item == elem)
            return true;
    });
    return false;
};


console.log(__in__(3,[1,2,3]));

You have, essentially, function(){ return true; } in both cases -- it returns from the anonymous function, not from the one that called it! — JJJ
– JJJ, Commented Dec 28, 2012 at 19:43
This is probably just an example function, but there's also $.inArray(). — JJJ
– JJJ, Commented Dec 28, 2012 at 19:46
@F.Hauri "Better to remain silent and be thought a fool than to speak out and remove all doubt in the end" - Licoln. Of course jQ is Javascript... — jeremy
– jeremy, Commented Dec 28, 2012 at 20:19

Alex · Accepted Answer · 2012-12-28 19:45:16Z

11

That is because $.each it is executing function callbacks. $.each is not a for loop. The following is what you're after:

var __in__ = function(elem, array){
    var result = false;

    $.each(array, function(index, item) {
        if (item == elem) {
            result = true;
            return false;
        }
    });

    return result;
};

http://jsfiddle.net/unByH/1/

edited Dec 28, 2012 at 19:45

answered Dec 28, 2012 at 19:39

Alex

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Brian FitzGerald · Accepted Answer · 2012-12-28 19:42:02Z

4

Your inner return statement only returns from the inner (nested) function, the outer return statement is always running (even on a match).

answered Dec 28, 2012 at 19:42

Brian FitzGerald

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Matteo Tassinari · Accepted Answer · 2012-12-28 19:53:22Z

3

Try this:

var __in__ = function(elem, array){
    var found = false;

    $.each(array, function(index, item) {
        if (item == elem) {
            found = true;
            return false; // break out of $.each call
        }
    });

    return found;
};

edited Dec 28, 2012 at 19:53

answered Dec 28, 2012 at 19:43

Matteo Tassinari

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1 Comment

Matteo Tassinari Over a year ago

When I started typing there was no complete answer yet. Also, it is not mandatory to return from the each call but I agree that, performance-wise, it should be done, so I'll fix it.

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