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.