Checking for undefined parameter in js function

Question

I need some condition to catch and throw error when a non-function data type is passed as the second argument.

Including undefined being passed?

function test(fn) {
  console.log(fn)
  // should throw an error if a non-function data type is 
  //passed as the second argument (includes undefined being passed)
  if (fn && typeof fn !== 'function') throw Error();
  return fn;
}


function demo() {}

test(); //undefined OK!

test(undefined); // SHOULD THROW ERROR ALSO

// throws error since argument is defined and not a function
test('sdfasfs');

Possible duplicate of stackoverflow.com/questions/5999998/… — Johannes
– Johannes, Commented May 11, 2018 at 4:58
So you want to be able to distinguish between test() and test(undefined)? Or should those be treated the same? — Felix Kling
– Felix Kling, Commented May 11, 2018 at 5:17

Felix Kling · Accepted Answer · 2018-05-11 05:52:12Z

1

The only way to distinguish between implicitly and explicitly passing undefined is to look at how many arguments have been passed:

function test(fn) {
  console.log(fn)
  // should throw an error if a non-function data type is 
  //passed as the second argument (includes undefined being passed)
  if (arguments.length > 0 && typeof fn !== 'function') throw Error();
  return fn;
}


function demo() {}

test(); //undefined OK!

test(undefined); // SHOULD THROW ERROR ALSO

// throws error since argument is defined and not a function
test('sdfasfs');

answered May 11, 2018 at 5:52

Felix Kling

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Comments

Nikhil Aggarwal · Accepted Answer · 2018-05-11 05:13:11Z

1

It will not throw an error because of fn condition in if which will be evaluated to false for undefined and hence, if is not evaluated to true and error is not thrown.

If you are looking for throwing an error if undefined is passed then, you can update your code to following

function test(fn) {
  console.log(fn)
  // should throw an error if a non-function data type is 
  //passed as the second argument (includes undefined being passed)
  if ((fn || fn === undefined) && typeof fn !== 'function') throw Error();
  return fn;
}

function demo() {}

test(); //undefined OK!

test(undefined); // WILL THROW ERROR

// throws error since argument is defined and not a function
test('sdfasfs');

edited May 11, 2018 at 5:13

answered May 11, 2018 at 5:00

Nikhil Aggarwal

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Comments

felixmosh · Accepted Answer · 2018-05-11 05:01:08Z

0

Because of the fact that undefined is falsy in JS, your first condition prevent you from throwing.

Just remove the first condition.

function test(fn) {
  console.log(fn)
  // should throw an error if a non-function data type is 
  //passed as the second argument (includes undefined being passed)
  if (typeof fn !== 'function') throw Error();
  return fn;
}


function demo() {}

test(); //undefined OK!

test(undefined); // SHOULD THROW ERROR ALSO

// throws error since argument is defined and not a function
test('sdfasfs');

answered May 11, 2018 at 5:01

felixmosh

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1 Comment

Felix Kling Over a year ago

Then test() will throw though. Maybe that's what the OP wants, but it's unclear to me because of the //undefined OK! comment.

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