typeof won't work when checking a function that's passed as a parameter

Question

New to javascript. I want to check if temp is a function or not. Also I would like to know why typeof won't work in this situation : Situation where a function is passed as a parameter. Understanding it is my purpose so no jQuery please. Appreciate all the help. Thanks

function getParams(foo, bar) {
  if (typeof bar === 'function') console.log("bar is a function");
  console.log(typeof bar); // string: because i returned string. But why not a "function" ? 
}

function temp(element) {
  return element;
}

function runThis() {
  getParams("hello", temp("world"));
}


runThis();

You're not passing a function, you're passing the results of calling a function. — Dave Newton, Oct 15 '19 at 16:36
@DaveNewton Is right, you're passing a returned value. Which is a string. — Josh, Oct 15 '19 at 16:37
Thanks Dave. Is there a way to pass a function purely, not the result of calling a function? — Newbie_Android, Oct 15 '19 at 16:37
@Newbie_Android Stop calling it--that's what the trailing `()` does like every other function you call. — Dave Newton, Oct 15 '19 at 16:39
To pass a function with parameters, pass an anonymous function that calls your function with the required parameter: `getParams("hello", function() { temp("world"); });`. — Racil Hilan, Oct 15 '19 at 17:07
@Racil Hilan Thank you so much, I think that was what i was looking for — Newbie_Android, Oct 15 '19 at 17:14

mwilson · Accepted Answer · 2019-10-15T16:47:43.440

temp('world') returns a string so therefore you are passing in a string instead of a function.

did you mean to pass in temp instead?

function getParams(foo, bar) {
  if (typeof bar === 'function') console.log("bar is a function");
  console.log(typeof bar); // string: because i returned string. But why not a "function" ? 
}

function temp(element) {
  return element;
}

function runThis() {
  getParams("hello", temp("world")); // <-- temp("world") isn't a function. It's the result of a function
}

// Did you mean to do this?
function runThis2() {
  getParams("hello", temp);
}


runThis();
runThis2();

If you are wanting to also pass parameters to your passed in function, you could do something like this (there are multiple ways to accomplish this):

function getParams(foo, bar, functionParam) {
  if (typeof bar === 'function') 
  {
    console.log("bar is a function");
    const result = bar(functionParam);
    console.log('function result: ', result);
  }
  console.log(typeof bar); // string: because i returned string. But why not a "function" ? 
}

function temp(element) {
  return element;
}

// Did you mean to do this?
function runThis2() {
  getParams("hello", temp, 'my function param');
}

runThis2();

ahh.. i see. Thanks. But I need to pass parameters to the temp function. As in, : getParams(a1, a2, temp(a1,a2,"value")) to check. a1 and a2 need to be passed to the function, "temp" as parameters. — Newbie_Android, Oct 15 '19 at 16:42
yea I will do that. Thanks. I had no idea I was passing a result of a function if i do that way. — Newbie_Android, Oct 15 '19 at 16:46

typeof won't work when checking a function that's passed as a parameter

1 Answers1