JavaScript function as function parameter default value

Question

What would be the standard way to define a function which has an optional argument, which itself is a function?

For example I want anotherFunction() to return true, if it's not defined.

function myFunction ( anotherFunction ) {
    /*
    some code here
    */
    return anotherFunction ();
}

see here: stackoverflow.com/questions/13286233/…

bovino Marcelo Bezerra
– bovino Marcelo Bezerra

2014-06-28 18:25:46 +00:00
Commented Jun 28, 2014 at 18:25 — bovino Marcelo Bezerra
– bovino Marcelo Bezerra, Commented Jun 28, 2014 at 18:25

Lucas Trzesniewski · Accepted Answer · 2014-06-28 18:30:42Z

2

function myFunction ( anotherFunction ) {
    /*
    some code here
    */
    return (typeof anotherFunction == "function") ? anotherFunction() : true;
}

This has the side effect of making sure the arguement is a function and not some garbage. If you'd prefer to throw, just use return anotherFunction ? anotherFunction() : true;.

edited Jun 28, 2014 at 18:30

answered Jun 28, 2014 at 18:25

Lucas Trzesniewski

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3 Comments

user229044 Over a year ago

The "additional benefit" isn't an additional benefit. If you function expects an argument which is a function, it should raise an error if a non-function is passed in. It's better to detect that and fail early, than for the function to happily proceed, disguising the bug.

Lucas Trzesniewski Over a year ago

That's true, but it entirely depends on what you're trying to achieve.

vahidseo Over a year ago

Thanks. I used typeof anotherFunction !== "undefined" instead.

user229044 · Accepted Answer · 2014-06-28 18:26:52Z

1

Just test if a value was passed:

return anotherFunction ? anotherFunction() : true

answered Jun 28, 2014 at 18:26

user229044♦

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Comments

Peter Herdenborg · Accepted Answer · 2014-06-28 18:29:23Z

1

I'd use

function myFunction ( anotherFunction ) {
    anotherFunction = anotherFunction || function(){ return true; };
    /*
    some code here
    */
    return anotherFunction ();
}

This corresponds to the OP's desire to default the optional parameter with a function, not merely making myFunction return true.

answered Jun 28, 2014 at 18:29

Peter Herdenborg

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3 Comments

user229044 Over a year ago

You're defining and invoking a function needlessly every time myFunction is called.

Lucas Trzesniewski Over a year ago

A workaround would be to define var trueFn = function() { return true; };, and then use anotherFunction = anotherFunction || trueFn;

Peter Herdenborg Over a year ago

True. It may not be the best idea to do this from a performance point of view but conceptually I think it's the most accurate approach. In the general case, the default function could be more meaningful than merely returning true.

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