Is it possible to send a variable number of arguments to a JavaScript function?

Is it possible to send a variable number of arguments to a JavaScript function, from an array?

var arr = ['a','b','c']

var func = function()
{
    // debug 
    alert(arguments.length);
    //
    for(arg in arguments)
        alert(arg);
}

func('a','b','c','d'); // prints 4 which is what I want, then 'a','b','c','d'
func(arr); // prints 1, then 'Array'

I've recently written a lot of Python and it's a wonderful pattern to be able to accept varargs and send them. e.g.

def func(*args):
   print len(args)
   for i in args:
       print i

func('a','b','c','d'); // prints 4 which is what I want, then 'a','b','c','d'
func(*arr) // prints 4 which is what I want, then 'a','b','c','d'

Is it possible in JavaScript to send an array to be treated as the arguments array?

Update: Since ES6, you can simply use the spread syntax when calling the function:

func(...arr);

Since ES6 also if you expect to treat your arguments as an array, you can also use the spread syntax in the parameter list, for example:

function func(...args) {
  args.forEach(arg => console.log(arg))
}

const values = ['a', 'b', 'c']
func(...values)
func(1, 2, 3)

And you can combine it with normal parameters, for example if you want to receive the first two arguments separately and the rest as an array:

function func(first, second, ...theRest) {
  //...
}

And maybe is useful to you, that you can know how many arguments a function expects:

var test = function (one, two, three) {}; 
test.length == 3;

But anyway you can pass an arbitrary number of arguments...

The spread syntax is shorter and "sweeter" than apply and if you don't need to set the this value in the function call, this is the way to go.

Here is an apply example, which was the former way to do it:

var arr = ['a','b','c'];

function func() {
  console.log(this); // 'test'
  console.log(arguments.length); // 3

  for(var i = 0; i < arguments.length; i++) {
    console.log(arguments[i]);
  }

};

func.apply('test', arr);

Nowadays I only recommend using apply only if you need to pass an arbitrary number of arguments from an array and set the this value. apply takes is the this value as the first arguments, which will be used on the function invocation, if we use null in non-strict code, the this keyword will refer to the Global object (window) inside func, in strict mode, when explicitly using 'use strict' or in ES modules, null will be used.

Also note that the arguments object is not really an Array, you can convert it by:

var argsArray = Array.prototype.slice.call(arguments);

And in ES6:

const argsArray = [...arguments] // or Array.from(arguments)

But you rarely use the arguments object directly nowadays thanks to the spread syntax.

You can actually pass as many values as you want to any javascript function. The explicitly named parameters will get the first few values, but ALL parameters will be stored in the arguments array.

To pass the arguments array in "unpacked" form, you can use apply, like so (c.f. Functional Javascript):

var otherFunc = function() {
   alert(arguments.length); // Outputs: 10
}

var myFunc = function() {
  alert(arguments.length); // Outputs: 10
  otherFunc.apply(this, arguments);
}
myFunc(1,2,3,4,5,6,7,8,9,10);

The splat and spread operators are part of ES6, the planned next version of Javascript. So far only Firefox supports them. This code works in FF16+:

var arr = ['quick', 'brown', 'lazy'];

var sprintf = function(str, ...args)
{
    for (arg of args) {
        str = str.replace(/%s/, arg);
    }
    return str;
}

sprintf.apply(null, ['The %s %s fox jumps over the %s dog.', ...arr]);
sprintf('The %s %s fox jumps over the %s dog.', 'slow', 'red', 'sleeping');

Note the awkard syntax for spread. The usual syntax of sprintf('The %s %s fox jumps over the %s dog.', ...arr); is not yet supported. You can find an ES6 compatibility table here.

Note also the use of for...of, another ES6 addition. Using for...in for arrays is a bad idea.

There are two methods as of ES2015.

The arguments Object

This object is built into functions, and it refers to, appropriately, the arguments of a function. It is not technically an array, though, so typical array operations won't work on it. The suggested method is to use Array.from or the spread operator to create an array from it.

I've seen other answers mention using slice. Don't do that. It prevents optimizations (source: MDN).

Array.from(arguments)
[...arguments]

However, I would argue that arguments is problematic because it hides what a function accepts as input. An arguments function typically is written like this:

function mean(){
    let args = [...arguments];
    return args.reduce((a,b)=>a+b) / args.length;
}

Sometimes, the function header is written like the following to document the arguments in a C-like fashion:

function mean(/* ... */){ ... }

But that is rare.

As for why it is problematic, take C, for example. C is backwards compatible with an ancient pre-ANSI dialect of the language known as K&R C. K&R C allows function prototypes to have an empty arguments list.

int myFunction();
/* This function accepts unspecified parameters */

ANSI C provides a facility for varargs (...), and void to specify an empty arguments-list.

int myFunction(void);
/* This function accepts no parameters */

Many people inadvertently declare functions with an unspecified arguments list (int myfunction();), when they expect the function to take zero arguments. This is technically a bug because the function will accept arguments. Any number of them.

A proper varargs function in C takes the form:

int myFunction(int nargs, ...);

And JavaScript actually does have something similar to this.

Spread Operator / Rest Parameters

I've already shown you the spread operator, actually.

...name

It's pretty versatile, and can also be used in a function's argument-list ("rest parameters") to specify varargs in a nicely documented fashion:

function mean(...args){
    return args.reduce((a,b)=>a+b) / args.length;
}

Or as a lambda expression:

((...args)=>args.reduce((a,b)=>a+b) / args.length)(1,2,3,4,5); // 3

I much prefer the spread operator. It is clean and self-documenting.