How to use optional arguments in node.js

gistfile1.js

// example function where arguments 2 and 3 are optional
function example( err, optionalA, optionalB, callback ) {

    // retrieve arguments as array
    var args = [];
    for (var i = 0; i < arguments.length; i++) {
        args.push(arguments[i]);
    }

    // first argument is the error object
    // shift() removes the first item from the
    // array and returns it
    err = args.shift();

    // last argument is the callback function.
    // pop() removes the last item in the array
    // and returns it
    callback = args.pop();

    // if args still holds items, these are
    // your optional items which you could
    // retrieve one by one like this:
    if (args.length > 0) optionalA = args.shift(); else optionalA = null;
    if (args.length > 0) optionalB = args.shift(); else optionalB = null;

    // continue as usual: check for errors
    if (err) return callback(err);

    // for tutorial purposes, log the optional parameters
    console.log('optionalA:', optionalA);
    console.log('optionalB:', optionalB);

    /* do your thing */

} // example()


// invoke example function with and without optional arguments

example(null, function (err) {   /* do something */    });

example(null, 'AA', function (err) {});

example(null, 'AAAA', 'BBBB', function (err) {});

mercy

Can we optimize the for loop by replacing it with Array.prototype.push.apply(args, arguments); ?

I had to remove undefined elements from the array as I kept on getting undefined is not a function error when trying to use the callback. Inserted this at line 9 and it seems to work a charm.

args = args.filter(function (val) {
  return val !== undefined;
});

@dandv Array.prototype.push.apply(args, arguments); works nicely for me

nice. thanks.

@dandv why not just var args = [].slice.call(arguments); ?

How do you know if optionalA or optionalB is intended?

If your pattern is function(required, optional, required) or function(required, optional, optional) I think it's OK to deal with those the way you have or any other way, but if you have a funcation signature like function(required, optional_1, optional_2, required) you have no way of telling what the user meant to send, function(required, optional_1, required) or function(required, optional_2, required)

Or have I missed something?

@adrianblynch That's only the case if optional_1 and optional_2 are independent. If you structure your inputs to function in such a way where optional_2 can be present only if optional_1 is present (even if that requires sending function(required, null, optional_2, required), you don't have to worry about that.

If the signature is function(required, optional_1, optional_2) where optional_1 and optional_2 are independent of each other and you pass in function(required, optional_2), unless you design your input to match specific unique criterion, you're right: there's no way to for the function know that you mean optional_2 instead of optional_1.

Long Story Short: design your function to require optional_1 in order to accept optional_2.

Cool. Thanks!

I get Error: SyntaxError: Unexpected string where the arguments is used. If i remove it, everything seems to work fine. Any help please? Thanks

thanks

Thank you @klovadis and @rossipedia !