Skip arguments in a JavaScript function

Question

I have a function like this:

function foo(a, b, c, d, e, f) { } 

In order to call this function only with an f argument, I know I should do:

foo(undefined, undefined, undefined, undefined, undefined, theFValue); 

Is there a less verbose way to do this?

Solutions:
I selected some proposed solutions (without using helper third functions)

// zero - ideal one, actually not possible(?!) foo(f: fValue);  // one - asks a "strange" declaration var _ = undefined; foo(_, _, _, _, _, fValue);  // two - asks the {} to be used instead of a 'natural' list of args //     - users should be aware about the internal structure of args obj //       so this option is not 'intellisense friendly' function foo(args){     // do stuff with `args.a`, `args.b`, etc. }     foo({f: fValue}); 

Answer 1

Such:

foo(undefined, undefined, undefined, undefined, undefined, arg1, arg2); 

.is equal to:

foo(...Array(5), arg1, arg2); 

.or:

foo(...[,,,,,], arg1, arg2); 

---

Such:

foo(undefined, arg1, arg2); 

.is equal to:

foo(...Array(1), arg1, arg2); 

.or:

foo(...[,], arg1, arg2); 

---

Such:

foo(arg1, arg2); 

.is equal to:

foo(...Array(0), arg1, arg2); 

.or:

foo(...[], arg1, arg2);