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Test 1:

var arr1 = [10, 20]; var arr2 = arr1; arr2[0]++; alert(arr1);// [11, 20] --- What?!?! alert(arr2);// [11, 20] --- correct

Test 2:

var arr1 = [10, 20]; var arr2 = [arr1[0], arr1[1]]; arr2[0]++; alert(arr1);// [10, 20] --- correct alert(arr2);// [11, 20] --- correct

In test 1, Why the first array's first element was changed? As far as I know, in other OOP languages like Java & PHP, if we do the Test 1's var arr2 = arr1; this is called "referencing", once we change something with the new variable's value, it creates a new value for that so the first reference "arr1" doesn't get effected. Why it's not the same thing in Javascript!? This totally doesn't make sense! Is it a bug? or it's just that Javascript behaves differently?

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    Look up "javascript deep copy" vs "javascript shallow copy" Commented Jun 29, 2014 at 0:25

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JavaScript variables refer to objects. When a variable "is" an object, it contains a reference to the object. Assigning the value of such a variable to another variable just transfers the reference; it does not make a copy of the referenced object.

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In the Test 1 both variables are referencing the same array object, so anything you do on one is going to be reflected on the other one.

On the other hand, on the Test 2 you are copying the values into a new array, so each variable is referencing different array objects.

edit: to remove the "passby" term.

edit 2: on a side note, this happens in java too.

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This is really not correct. There's no "passing" happening here. The terms "pass by value" and "pass by reference" have to do with the semantics of function calls.
As a side-note, an easier way to copy the values is arr2 = arr1.slice(0);.
lol, true passing means as parameters in a function/method, but its the same idea.
@EdwardM.B. it's a related concept, but strictly speaking they're really two completely different things.
Lots of languages do this. The big exceptions are truly functional languages where data's immutable.

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