Question is on JavaScript function. Can anybody answer this.

Question

//case 1)
function hello(x)
{
alert("hello" + x);
}
hello(3);
//output: hello3

//Case 2)
function hello(x)
{
alert("hello" + x);
}
hello()
//Output: helloundefined.

case 3)
function hello()
{
alert("hello" + x)
}
hello(3);
//Output: error

//case4)
function hello()
{
alert("hello" + 3);
}
hello(3);
//Output: hello3

//Why in 4th case it gives output unlike 3rd case.

Answer 1

Javascript does not care about function signature, you can pass any number of arguments to a function even if they are not in the function definition. they will just be undefined.

Answer 2

look at the alerts of case 3 and case 4

Answer 3

Because you don't have an undefined variable being passed to the alert function. That's why you don't have an error.

Answer 4

Because in 3rd you are giving a value to parameter which is not declared in your function ,while in 4th try only
alert("hello"+3);
Without declaring a function u will get hello3 as o/p

Answer 5

case 3 // x has to be defined somewhere since it's not passed to the function.

Answer 6

case 3 alert depended on param x (x not defined)

case 4 alert not depended on any param

Answer 7

ODLNT I know .But I want reason.. why it is giving output in 4th case

Answer 8

ODLNT Bahha🐧 Okay understood.
so ,if function doesn't have any argument, calling function may contain argument and gives result only if [alert box or codes]contains undefined variable.?

I thought it shows error .