Javascript returns string for OR(||) operation

Question

I am unable to understand this.

Following is expression uses OR operator

var subCond1 = adj.getData(\'relationEnabled\') == \'true\' || adj.getData(\'unh         


        

Answer 1

Yup, that's just one of the features of || in JavaScript, and it's deliberate. It doesn't return a boolean (necessarily), it works like this: It evaluates the left-hand operand and if that operand is truthy, it returns it; otherwise, it evalutes and returns the right-hand operand.

So what's "truthy"? Anything that isn't "falsy". :-) The falsy values are 0, "", null, undefined, NaN, and of course, false. Anything else is truthy.

If you need a boolean, just !! the result:

var subCond1 = !!(adj.getData('relationEnabled') == 'true' || adj.getData('unhideIfHidden') || adj.getData('hlFixed') == 'true');

...but you frequently don't need to bother.

This behavior of || is really useful, particularly (I find) when dealing with object references that may be null, when you want to provide a default:

var obj = thisMayBeNull || {};

Now, obj will be thisMayBeNull if it's truthy (non-null object references are truthy), or {} if thisMayBeNull is falsy.

More in this article on my blog: JavaScript's Curiously-Powerful OR Operator (||)

Just to round things out: The && operator has a similar behavior: It evaluates the left-hand operand and, if it's falsy, returns it; otherwise it evaluates and returns the right-hand operator. This is useful if you want an object property from a variable which may be null:

var value = obj && obj.property;

value will be the value of obj if obj is falsy (for instance, null), or the value of obj.property if obj is truthy.