Is there a better way to do optional function parameters in JavaScript?

Question

I've always handled optional parameters in JavaScript like this:

function myFunc(requiredArg, optionalArg){
  optionalArg = optionalArg || 'defaultValue';

  // Do stuff
}

Is there a better way to do it?

Are there any cases where using || like that is going to fail?

Answer 1

Your logic fails if optionalArg is passed, but evaluates as false - try this as an alternative

if (typeof optionalArg === 'undefined') { optionalArg = 'default'; }

Or an alternative idiom:

optionalArg = (typeof optionalArg === 'undefined') ? 'default' : optionalArg;

Use whichever idiom communicates the intent best to you!

Answer 2

In ECMAScript 2015 (aka "ES6") you can declare default argument values in the function declaration:

function myFunc(requiredArg, optionalArg = 'defaultValue') {
    // do stuff
}

More about them in this article on MDN.

This is currently only supported by Firefox, but as the standard has been completed, expect support to improve rapidly.

---

EDIT (2019-06-12):

Default parameters are now widely supported by modern browsers.
All versions of Internet Explorer do not support this feature. However, Chrome, Firefox, and Edge currently support it.

Answer 3

I find this to be the simplest, most readable way:

if (typeof myVariable === 'undefined') { myVariable = 'default'; }
//use myVariable here

Paul Dixon's answer (in my humble opinion) is less readable than this, but it comes down to preference.

insin's answer is much more advanced, but much more useful for big functions!

EDIT 11/17/2013 9:33pm: I've created a package for Node.js that makes it easier to "overload" functions (methods) called parametric.

Answer 4

If you need to chuck a literal NULL in, then you could have some issues. Apart from that, no, I think you're probably on the right track.

The other method some people choose is taking an assoc array of variables iterating through the argument list. It looks a bit neater but I imagine it's a little (very little) bit more process/memory intensive.

function myFunction (argArray) {
    var defaults = {
        'arg1'  :   "value 1",
        'arg2'  :   "value 2",
        'arg3'  :   "value 3",
        'arg4'  :   "value 4"
    }

    for(var i in defaults) 
        if(typeof argArray[i] == "undefined") 
               argArray[i] = defaults[i];

    // ...
}

Answer 5

Ideally, you would refactor to pass an object and merge it with a default object, so the order in which arguments are passed doesn't matter (see the second section of this answer, below).

If, however, you just want something quick, reliable, easy to use and not bulky, try this:

---

A clean quick fix for any number of default arguments

• It scales elegantly: minimal extra code for each new default
• You can paste it anywhere: just change the number of required args and variables
• If you want to pass undefined to an argument with a default value, this way, the variable is set as undefined. Most other options on this page would replace undefined with the default value.

Here's an example for providing defaults for three optional arguments (with two required arguments)

function myFunc( requiredA, requiredB,  optionalA, optionalB, optionalC ) {

  switch (arguments.length - 2) { // 2 is the number of required arguments
    case 0:  optionalA = 'Some default';
    case 1:  optionalB = 'Another default';
    case 2:  optionalC = 'Some other default';
    // no breaks between cases: each case implies the next cases are also needed
  }

}

Simple demo. This is similar to roenving's answer, but easily extendible for any number of default arguments, easier to update, and using arguments not Function.arguments.

---

Passing and merging objects for more flexibility

The above code, like many ways of doing default arguments, can't pass arguments out of sequence, e.g., passing optionalC but leaving optionalB to fall back to its default.

A good option for that is to pass objects and merge with a default object. This is also good for maintainability (just take care to keep your code readable, so future collaborators won't be left guessing about the possible contents of the objects you pass around).

Example using jQuery. If you don't use jQuery, you could instead use Underscore's _.defaults(object, defaults) or browse these options:

function myFunc( args ) {
  var defaults = {
    optionalA: 'Some default',
    optionalB: 'Another default',
    optionalC: 'Some other default'
  };
  args = $.extend({}, defaults, args);
}

Here's a simple example of it in action.

Answer 6

You can use some different schemes for that. I've always tested for arguments.length:

function myFunc(requiredArg, optionalArg){
  optionalArg = myFunc.arguments.length<2 ? 'defaultValue' : optionalArg;

  ...

-- doing so, it can't possibly fail, but I don't know if your way has any chance of failing, just now I can't think up a scenario, where it actually would fail ...

And then Paul provided one failing scenario !-)

Answer 7

Similar to Oli's answer, I use an argument Object and an Object which defines the default values. With a little bit of sugar...

/**
 * Updates an object's properties with other objects' properties. All
 * additional non-falsy arguments will have their properties copied to the
 * destination object, in the order given.
 */
function extend(dest) {
  for (var i = 1, l = arguments.length; i < l; i++) {
    var src = arguments[i]
    if (!src) {
      continue
    }
    for (var property in src) {
      if (src.hasOwnProperty(property)) {
        dest[property] = src[property]
      }
    }
  }
  return dest
}

/**
 * Inherit another function's prototype without invoking the function.
 */
function inherits(child, parent) {
  var F = function() {}
  F.prototype = parent.prototype
  child.prototype = new F()
  child.prototype.constructor = child
  return child
}

...this can be made a bit nicer.

function Field(kwargs) {
  kwargs = extend({
    required: true, widget: null, label: null, initial: null,
    helpText: null, errorMessages: null
  }, kwargs)
  this.required = kwargs.required
  this.label = kwargs.label
  this.initial = kwargs.initial
  // ...and so on...
}

function CharField(kwargs) {
  kwargs = extend({
    maxLength: null, minLength: null
  }, kwargs)
  this.maxLength = kwargs.maxLength
  this.minLength = kwargs.minLength
  Field.call(this, kwargs)
}
inherits(CharField, Field)

What's nice about this method?

• You can omit as many arguments as you like - if you only want to override the value of one argument, you can just provide that argument, instead of having to explicitly pass undefined when, say there are 5 arguments and you only want to customise the last one, as you would have to do with some of the other methods suggested.
• When working with a constructor Function for an object which inherits from another, it's easy to accept any arguments which are required by the constructor of the Object you're inheriting from, as you don't have to name those arguments in your constructor signature, or even provide your own defaults (let the parent Object's constructor do that for you, as seen above when CharField calls Field's constructor).
• Child objects in inheritance hierarchies can customise arguments for their parent constructor as they see fit, enforcing their own default values or ensuring that a certain value will always be used.

Answer 8

Loose type checking

Easy to write, but 0, '', false, null and undefined will be converted to default value, which might not be expected outcome.

function myFunc(requiredArg, optionalArg) {
    optionalArg = optionalArg || 'defaultValue';
}

Strict type checking

Longer, but covers majority of cases. Only case where it incorrectly assigns default value is when we pass undefined as parameter.

function myFunc(requiredArg, optionalArg) {
    optionalArg = typeof optionalArg !== 'undefined' ? optionalArg : 'defaultValue';
}

Checking arguments variable

Catches all cases but is the most clumsy to write.

function myFunc(requiredArg, optionalArg1, optionalArg2) {
    optionalArg1 = arguments.length > 1 ? optionalArg1 : 'defaultValue';
    optionalArg2 = arguments.length > 2 ? optionalArg2 : 'defaultValue';
}

ES6

Unfortunately this has very poor browser support at the moment

function myFunc(requiredArg, optionalArg = 'defaultValue') {

}

Answer 9

If you're using defaults extensively, this seems much more readable:

function usageExemple(a,b,c,d){
    //defaults
    a=defaultValue(a,1);
    b=defaultValue(b,2);
    c=defaultValue(c,4);
    d=defaultValue(d,8);

    var x = a+b+c+d;
    return x;
}

Just declare this function on the global escope.

function defaultValue(variable,defaultValue){
    return(typeof variable!=='undefined')?(variable):(defaultValue);
}

Usage pattern fruit = defaultValue(fruit,'Apple');

*PS you can rename the defaultValue function to a short name, just don't use default it's a reserved word in javascript.

Answer 10

I am used to seeing a few basic variations on handling optional variables. Sometimes, the relaxed versions are useful.

function foo(a, b, c) {
  a = a || "default";   // Matches 0, "", null, undefined, NaN, false.
  a || (a = "default"); // Matches 0, "", null, undefined, NaN, false.

  if (b == null) { b = "default"; } // Matches null, undefined.

  if (typeof c === "undefined") { c = "default"; } // Matches undefined.
}

The falsy default used with variable a is, for example, used extensively in Backbone.js.

Answer 11

With ES2015/ES6 you can take advantage of Object.assign which can replace $.extend() or _.defaults()

function myFunc(requiredArg, options = {}) {
  const defaults = {
    message: 'Hello',
    color: 'red',
    importance: 1
  };

  const settings = Object.assign({}, defaults, options);

  // do stuff
}

You can also use defaults arguments like this

function myFunc(requiredArg, { message: 'Hello', color: 'red', importance: 1 } = {}) {
  // do stuff
}

Answer 12

If you're using the Underscore library (you should, it's an awesome library):

_.defaults(optionalArg, 'defaultValue');

Answer 13

I don't know why @Paul's reply is downvoted, but the validation against null is a good choice. Maybe a more affirmative example will make better sense:

In JavaScript a missed parameter is like a declared variable that is not initialized (just var a1;). And the equality operator converts the undefined to null, so this works great with both value types and objects, and this is how CoffeeScript handles optional parameters.

function overLoad(p1){
    alert(p1 == null); // Caution, don't use the strict comparison: === won't work.
    alert(typeof p1 === 'undefined');
}

overLoad(); // true, true
overLoad(undefined); // true, true. Yes, undefined is treated as null for equality operator.
overLoad(10); // false, false


function overLoad(p1){
    if (p1 == null) p1 = 'default value goes here...';
    //...
}

Though, there are concerns that for the best semantics is typeof variable === 'undefined' is slightly better. I'm not about to defend this since it's the matter of the underlying API how a function is implemented; it should not interest the API user.

I should also add that here's the only way to physically make sure any argument were missed, using the in operator which unfortunately won't work with the parameter names so have to pass an index of the arguments.

function foo(a, b) {
    // Both a and b will evaluate to undefined when used in an expression
    alert(a); // undefined
    alert(b); // undefined

    alert("0" in arguments); // true
    alert("1" in arguments); // false
}

foo (undefined);

Answer 14

The test for undefined is unnecessary and isn't as robust as it could be because, as user568458 pointed out, the solution provided fails if null or false is passed. Users of your API might think false or null would force the method to avoid that parameter.

function PaulDixonSolution(required, optionalArg){
   optionalArg = (typeof optionalArg === "undefined") ? "defaultValue" : optionalArg;
   console.log(optionalArg);
};
PaulDixonSolution("required");
PaulDixonSolution("required", "provided");
PaulDixonSolution("required", null);
PaulDixonSolution("required", false);

The result is:

defaultValue
provided
null
false

Those last two are potentially bad. Instead try:

function bulletproof(required, optionalArg){
   optionalArg = optionalArg ? optionalArg : "defaultValue";;
   console.log(optionalArg);
};
bulletproof("required");
bulletproof("required", "provided");
bulletproof("required", null);
bulletproof("required", false);

Which results in:

defaultValue
provided
defaultValue
defaultValue

The only scenario where this isn't optimal is when you actually have optional parameters that are meant to be booleans or intentional null.

Answer 15

I tried some options mentioned in here and performance tested them. At this moment the logicalor seems to be the fastest. Although this is subject of change over time (different JavaScript engine versions).

These are my results (Microsoft Edge 20.10240.16384.0):

Function executed            Operations/sec     Statistics
TypeofFunction('test');          92,169,505     ±1.55%   9% slower
SwitchFuntion('test');            2,904,685     ±2.91%  97% slower
ObjectFunction({param1: 'test'});   924,753     ±1.71%  99% slower
LogicalOrFunction('test');      101,205,173     ±0.92%     fastest
TypeofFunction2('test');         35,636,836     ±0.59%  65% slower

This performance test can be easily replicated on: http://jsperf.com/optional-parameters-typeof-vs-switch/2

This is the code of the test:

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script>
<script>
    Benchmark.prototype.setup = function() {
        function TypeofFunction(param1, optParam1, optParam2, optParam3) {
            optParam1 = (typeof optParam1 === "undefined") ? "Some default" : optParam1;
            optParam2 = (typeof optParam2 === "undefined") ? "Another default" : optParam2;
            optParam3 = (typeof optParam3 === "undefined") ? "Some other default" : optParam3;
        }

        function TypeofFunction2(param1, optParam1, optParam2, optParam3) {
            optParam1 = defaultValue(optParam1, "Some default");
            optParam2 = defaultValue(optParam2, "Another default");
            optParam3 = defaultValue(optParam3, "Some other default");
        }

        function defaultValue(variable, defaultValue) {
            return (typeof variable !== 'undefined') ? (variable) : (defaultValue);
        }

        function SwitchFuntion(param1, optParam1, optParam2, optParam3) {
            switch (arguments.length - 1) { // <-- 1 is number of required arguments
                case 0:
                    optParam1 = 'Some default';
                case 1:
                    optParam2 = 'Another default';
                case 2:
                    optParam3 = 'Some other default';
            }
        }

        function ObjectFunction(args) {
            var defaults = {
                optParam1: 'Some default',
                optParam2: 'Another default',
                optParam3: 'Some other default'
            }
            args = $.extend({}, defaults, args);
        }

        function LogicalOrFunction(param1, optParam1, optParam2, optParam3) {
            optParam1 || (optParam1 = 'Some default');
            optParam2 || (optParam1 = 'Another default');
            optParam3 || (optParam1 = 'Some other default');
        }
    };
</script>

Answer 16

Landed to this question, searching for default parameters in EcmaScript 2015, thus just mentioning...

With ES6 we can do default parameters:

function doSomething(optionalParam = "defaultValue"){
    console.log(optionalParam);//not required to check for falsy values
}

doSomething(); //"defaultValue"
doSomething("myvalue"); //"myvalue"

Answer 17

During a project I noticed I was repeating myself too much with the optional parameters and settings, so I made a class that handles the type checking and assigns a default value which results in neat and readable code. See example and let me know if this works for you.

var myCar           = new Car('VW', {gearbox:'automatic', options:['radio', 'airbags 2x']});
var myOtherCar      = new Car('Toyota');

function Car(brand, settings) {
    this.brand      = brand;

    // readable and adjustable code
    settings        = DefaultValue.object(settings, {});
    this.wheels     = DefaultValue.number(settings.wheels, 4);
    this.hasBreaks  = DefaultValue.bool(settings.hasBreaks, true);
    this.gearbox    = DefaultValue.string(settings.gearbox, 'manual');
    this.options    = DefaultValue.array(settings.options, []);

    // instead of doing this the hard way
    settings        = settings || {};
    this.wheels     = (!isNaN(settings.wheels)) ? settings.wheels : 4;
    this.hasBreaks  = (typeof settings.hasBreaks !== 'undefined') ? (settings.hasBreaks === true) : true;
    this.gearbox    = (typeof settings.gearbox === 'string') ? settings.gearbox : 'manual';
    this.options    = (typeof settings.options !== 'undefined' && Array.isArray(settings.options)) ? settings.options : [];
}

Using this class:

(function(ns) {

    var DefaultValue = {

        object: function(input, defaultValue) {
            if (typeof defaultValue !== 'object') throw new Error('invalid defaultValue type');
            return (typeof input !== 'undefined') ? input : defaultValue;
        },

        bool: function(input, defaultValue) {
            if (typeof defaultValue !== 'boolean') throw new Error('invalid defaultValue type');
            return (typeof input !== 'undefined') ? (input === true) : defaultValue;
        },

        number: function(input, defaultValue) {
            if (isNaN(defaultValue)) throw new Error('invalid defaultValue type');
            return (typeof input !== 'undefined' && !isNaN(input)) ? parseFloat(input) : defaultValue;
        },

        // wrap the input in an array if it is not undefined and not an array, for your convenience
        array: function(input, defaultValue) {
            if (typeof defaultValue === 'undefined') throw new Error('invalid defaultValue type');
            return (typeof input !== 'undefined') ? (Array.isArray(input) ? input : [input]) : defaultValue;
        },

        string: function(input, defaultValue) {
            if (typeof defaultValue !== 'string') throw new Error('invalid defaultValue type');
            return (typeof input === 'string') ? input : defaultValue;
        },

    };

    ns.DefaultValue = DefaultValue;

}(this));

Answer 18

This is what I ended up with:

function WhoLikesCake(options) {
  options = options || {};
  var defaultOptions = {
    a : options.a || "Huh?",
    b : options.b || "I don't like cake."
  }
  console.log('a: ' + defaultOptions.b + ' - b: ' + defaultOptions.b);

  // Do more stuff here ...
}

Called like this:

WhoLikesCake({ b : "I do" });

Answer 19

Folks -

After looking at these and other solutions, I tried a number of them out using a snippet of code originally from W3Schools as a base. You can find what works in the following. Each of the items commented out work as well and are that way to allow you to experiment simply by removing individual comments. To be clear, it is the "eyecolor" parameter that is not being defined.

function person(firstname, lastname, age, eyecolor)
{
this.firstname = firstname;
this.lastname = lastname;
this.age = age;
this.eyecolor = eyecolor;
// if(null==eyecolor)
//   this.eyecolor = "unknown1";
//if(typeof(eyecolor)==='undefined') 
//   this.eyecolor = "unknown2";
// if(!eyecolor)
//   this.eyecolor = "unknown3";
this.eyecolor = this.eyecolor || "unknown4";
}

var myFather = new person("John", "Doe", 60);
var myMother = new person("Sally", "Rally", 48, "green");

var elem = document.getElementById("demo");
elem.innerHTML = "My father " +
              myFather.firstname + " " +
              myFather.lastname + " is " +
              myFather.age + " with " +
              myFather.eyecolor + " eyes.<br/>" +
              "My mother " +
              myMother.firstname + " " +
              myMother.lastname + " is " +
              myMother.age + " with " +
              myMother.eyecolor + " eyes."; 

Answer 20

function Default(variable, new_value)
{
    if(new_value === undefined) { return (variable === undefined) ? null : variable; }
    return (variable === undefined) ? new_value : variable;
}

var a = 2, b = "hello", c = true, d;

var test = Default(a, 0),
test2 = Default(b, "Hi"),
test3 = Default(c, false),
test4 = Default(d, "Hello world");

window.alert(test + "\n" + test2 + "\n" + test3 + "\n" + test4);

http://jsfiddle.net/mq60hqrf/

Answer 21

Here is my solution. With this you can leave any parameter you want. The order of the optional parameters is not important and you can add custom validation.

function YourFunction(optionalArguments) {
            //var scope = this;

            //set the defaults
            var _value1 = 'defaultValue1';
            var _value2 = 'defaultValue2';
            var _value3 = null;
            var _value4 = false;

            //check the optional arguments if they are set to override defaults...
            if (typeof optionalArguments !== 'undefined') {

                if (typeof optionalArguments.param1 !== 'undefined')
                    _value1 = optionalArguments.param1;

                if (typeof optionalArguments.param2 !== 'undefined')
                    _value2 = optionalArguments.param2;

                if (typeof optionalArguments.param3 !== 'undefined')
                    _value3 = optionalArguments.param3;

                if (typeof optionalArguments.param4 !== 'undefined')
                    //use custom parameter validation if needed, in this case for javascript boolean
                   _value4 = (optionalArguments.param4 === true || optionalArguments.param4 === 'true');
            }

            console.log('value summary of function call:');
            console.log('value1: ' + _value1);
            console.log('value2: ' + _value2);
            console.log('value3: ' + _value3);
            console.log('value4: ' + _value4);
            console.log('');
        }


        //call your function in any way you want. You can leave parameters. Order is not important. Here some examples:
        YourFunction({
            param1: 'yourGivenValue1',
            param2: 'yourGivenValue2',
            param3: 'yourGivenValue3',
            param4: true,
        });

        //order is not important
        YourFunction({
            param4: false,
            param1: 'yourGivenValue1',
            param2: 'yourGivenValue2',
        });

        //uses all default values
        YourFunction();

        //keeps value4 false, because not a valid value is given
        YourFunction({
            param4: 'not a valid bool'
        });

Answer 22

1. arg || 'default' is a great way and works for 90% of cases

2. It fails when you need to pass values that might be 'falsy'

   • false
   • 0
   • NaN
   • ""

   For these cases you will need to be a bit more verbose and check for undefined

3. Also be careful when you have optional arguments first, you have to be aware of the types of all of your arguments

Answer 23

In all cases where optionalArg is falsy you will end up with defaultValue.

function myFunc(requiredArg, optionalArg) {
    optionalArg = optionalArg || 'defaultValue';
    console.log(optionalArg);
    // Do stuff
}
myFunc(requiredArg);
myFunc(requiredArg, null);
myFunc(requiredArg, undefined);
myFunc(requiredArg, "");
myFunc(requiredArg, 0);
myFunc(requiredArg, false);

All of the above log defaultValue, because all of 6 are falsy. In case 4, 5, 6 you might not be interested to set optionalArg as defaultValue, but it sets since they are falsy.

Answer 24

Correct me if I'm wrong, but this seems like the simplest way (for one argument, anyway):

function myFunction(Required,Optional)
{
    if (arguments.length<2) Optional = "Default";
    //Your code
}

Answer 25

Those ones are shorter than the typeof operator version.

function foo(a, b) {
    a !== undefined || (a = 'defaultA');
    if(b === undefined) b = 'defaultB';
    ...
}

Answer 26

I suggest you to use ArgueJS this way:

function myFunc(){
  arguments = __({requiredArg: undefined, optionalArg: [undefined: 'defaultValue'})

  //do stuff, using arguments.requiredArg and arguments.optionalArg
  //    to access your arguments

}

You can also replace undefined by the type of the argument you expect to receive, like this:

function myFunc(){
  arguments = __({requiredArg: Number, optionalArg: [String: 'defaultValue'})

  //do stuff, using arguments.requiredArg and arguments.optionalArg
  //    to access your arguments

}

Answer 27

It seems that the safest way - to deal with all \ any falsy types of supplied arguments before deciding to use the default - is to check for the existence\presence of the optional argument in the invoked function.

Relying on the arguments object member creation which doesn't even get created if the argument is missing, regardless of the fact that it might be declared, we can write your function like this:

  function myFunc(requiredArg, optionalArg){
        optionalArg = 1 in arguments ? optionalArg : 'defaultValue';
  //do stuff
  }

Utilizing this behavior: We can safely check for any missing values on arguments list arbitrarily and explicitly whenever we need to make sure the function gets a certain value required in its procedure.

In the following demo code we will deliberately put a typeless and valueless undefined as a default value to be able to determine whether it might fail on falsy argument values, such as 0 false etc., or if it behaves as expected.

function argCheck( arg1, arg2, arg3 ){

       arg1 = 0 in arguments || undefined;
       arg2 = 1 in arguments || false;
       arg3 = 2 in arguments || 0;
   var arg4 = 3 in arguments || null;

   console.log( arg1, arg2, arg3, arg4 ) 
}

Now, checking few falsy argument-values to see if their presence is correctly detected and therefore evaluates to true:

argCheck( "", 0, false, null );
>> true true true true

Which means -they didn't fail the recognition of/as expected argument values. Here we have a check with all arguments missing, which according to our algo should acquire their default values even if they're falsy.

argCheck( );
>> undefined false 0 null

As we can see, the arguments arg1, arg2, arg3 and the undeclared arg4, are returning their exact default values, as ordered. Because we've now made sure that it works, we can rewrite the function which will actually be able to use them as in the first example by using: either if or a ternary condition.

On functions that have more than one optional argument, - a loop through, might have saved us some bits. But since argument names don't get initialized if their values are not supplied, we cannot access them by names anymore even if we've programmatically written a default value, we can only access them by arguments[index] which useless code readability wise.

But aside from this inconvenience, which in certain coding situations might be fully acceptable, there's another unaccounted problem for multiple and arbitrary number of argument defaults. Which may and should be considered a bug, as we can no longer skip arguments, as we once might have been able to, without giving a value, in a syntax such as:

argCheck("a",,22,{});

because it will throw! Which makes it impossible for us to substitute our argument with a specific falsy type of our desired default value. Which is stupid, since the arguments object is an array-like object and is expected to support this syntax and convention as is, natively or by default!

Because of this shortsighted decision we can no longer hope to write a function like this:

function argCheck( ) {
    var _default = [undefined, 0, false, null ],
        _arg = arguments;

 for( var x in _default ) {
         x in _arg ? 1 : _arg[x] = _default[x];
        }
    console.log( _arg[0],_arg[1],_arg[2],_arg[3] );
}

in which case, we would be able to write each default value of a desired type in arguments row and be able to at least access them by args.index.

For instance this function call would yield:

argCheck();
>>undefined 0 false null

as defined in our default array of arguments values. However the following is still possible:

argCheck({})
>>Object {  } 0 false null

argCheck({}, [])
>>Object {  } Array [  ] false null

But regretfully not:

 argCheck("a",,,22);
 >>SyntaxError: expected expression, got ','

Which would otherwise be logging:

>>a 0 false 22

but that's in a better world! However - for the original question - the topmost function will do just fine. e.g.:

function argCheck( arg, opt ) {
         1 in arguments ? 1 : opt = "default";
         console.log( arg, opt );
}

p.s.: sorry for not preserving the types of chosen defaults in my argument inputs while writing them.

Answer 28

function foo(requiredArg){
  if(arguments.length>1) var optionalArg = arguments[1];
}