Is there a better way to do optional function para

Question

This question already has an answer here:

• Set a default parameter value for a JavaScript function 21 answers

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

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 2

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 3

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 4

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 5

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 6

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 (despite the article title, they're called "arguments," not "parameters," in JavaScript).

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