if([]==true) //evalutes as false
//when i check empty array with true, if evaluating [] as false so it if condition return false
if([]) //evalutes as true
//when i check empty array alone, if evaluating [] as true so it if condition return true
why it is evaluating like that?
thanks
Based on abstract equality comparison algorithm your first code will be evaluated like below,
step 1 : ToNumber([]) == true
step 2 : ToPrimitive([]) == true
(ToNumber() will call ToPrimitve() when the passed argument is an object)
step 3 : "" == true
step 4 : 0 == true
step 5 : false == true
step 6 : false
And in your second case, [] is a truthy value, so if([]) will be evaluated to true always, here [] will not be converted as a primitive. Abstract equality comparison algorithm comes into play when you use == operator.
Another better example would be,
var x = [] || "hello";
console.log(x); // []
Since [] is a truthy value, x would be set with [] not "hello"
When you use only a variable as the condition (without comparison operators), Javascript will cast it to Boolean using the Boolean() function:
http://www.w3schools.com/js/js_booleans.asp
In your case, Boolean([]) = true so it returned as true.
