But in Java (and similar), you have to be explicit about varargs. In JavaScript, every function supports varargs. This will run just fine:

    function print(arg) {
        console.log(arg);
    }
    print("a", "b", "c");

It’ll only output "a", obviously. In fact, this will run just fine too:

    print();

You’ll get "undefined" in the console, but it’ll work.

JavaScript’s nature is that arguments are in the "arguments" pseudo-array, and the parameters are just fancy names for different values in "arguments". See:

    function count() {
        console.log(arguments.length);
    }
    count();
    count("a");
    count("a", "b");

In order, you’ll get: 0, 1, and 2. Despite the fact that "count" has no parameters in the definition.

In the first function ("print"), "arg" is just syntax sugar for "arguments[0]".

What I’m getting at is: in C, Java, etc., the compiler knows what is a varargs function and what isn’t. In JavaScript, the interpreter/compiler doesn’t and has to assume everything is.

Note that there's no such thing as spreaded arrays in Java. Instead, varargs are just syntactic sugar for arrays.

> Java Varargs functions can take spreaded arrays as Var parameters

That's syntactic sugar for an array, in the most literal way possible: when you define `func(Object… args)` the actual bytecode is that of `func(Object[] args)`, and you can pass in an array to stand in for the entire varargs.

So no, they're right. The varags just counts as 1 formal parameter.