Check if a variable is of function type

Question

Suppose I have any variable, which is defined as follows:

var a = function() {/* Statements */};

I want a function which checks if the type of the variable is function-like. i.e. :

function foo(v) {if (v is function type?) {/* do something */}};
foo(a);

How can I check if the variable a is of type Function in the way defined above?

Answer 1

Sure underscore's way is more efficient, but the best way to check, when efficiency isn't an issue, is written on underscore's page linked by @Paul Rosania.

Inspired by underscore, the final isFunction function is as follows:

function isFunction(functionToCheck) {
 return functionToCheck && {}.toString.call(functionToCheck) === '[object Function]';
}

Note: This solution doesn't work for async functions, generators or proxied functions. Please see other answers for more up to date solutions.

Answer 2

if (typeof v === 'function') {
    // do something
}

Answer 3

Underscore.js uses a more elaborate but highly performant test:

_.isFunction = function(obj) {
  return !!(obj && obj.constructor && obj.call && obj.apply);
};

See: http://jsperf.com/alternative-isfunction-implementations

EDIT: updated tests suggest that typeof might be faster, see http://jsperf.com/alternative-isfunction-implementations/4

Answer 4

There are several ways so I will summarize them all

1. Best way is:

   
   function foo(v) {if (v instanceof Function) {/* do something */} };
   
   

   Most performant (no string comparison) and elegant solution - the instanceof operator has been supported in browsers for a very long time, so don't worry - it will work in IE 6.
2. Next best way is:

   
   function foo(v) {if (typeof v === "function") {/* do something */} };
   
   

   disadvantage of typeof is that it is susceptible to silent failure, bad, so if you have a typo (e.g. "finction") - in this case the `if` will just return false and you won't know you have an error until later in your code
3. The next best way is:

   
   function isFunction(functionToCheck) {
       var getType = {};
       return functionToCheck && getType.toString.call(functionToCheck) === '[object Function]';
   }
   
   

   This has no advantage over solution #1 or #2 but is a lot less readable. An improved version of this is

   
   function isFunction(x) {
       return Object.prototype.toString.call(x) == '[object Function]';
   }
   
   

   but still lot less semantic than solution #1

Answer 5

jQuery (deprecated since version 3.3) Reference

$.isFunction(functionName);

AngularJS Reference

angular.isFunction(value);

Lodash Reference

_.isFunction(value);

Underscore Reference

_.isFunction(object); 

Node.js deprecated since v4.0.0 Reference

var util = require('util');
util.isFunction(object);

Answer 6

@grandecomplex: There's a fair amount of verbosity to your solution. It would be much clearer if written like this:

function isFunction(x) {
  return Object.prototype.toString.call(x) == '[object Function]';
}

Answer 7

var foo = function(){};
if (typeof foo === "function") {
  alert("is function")
}

Answer 8

Try the instanceof operator: it seems that all functions inherit from the Function class:

// Test data
var f1 = function () { alert("test"); }
var o1 = { Name: "Object_1" };
F_est = function () { };
var o2 = new F_est();

// Results
alert(f1 instanceof Function); // true
alert(o1 instanceof Function); // false
alert(o2 instanceof Function); // false

Answer 9

Something with more browser support and also include async functions could be:

const isFunction = value => value && (Object.prototype.toString.call(value) === "[object Function]" || "function" === typeof value || value instanceof Function);

and then test it like:

isFunction(isFunction); //true
isFunction(function(){}); //true
isFunction(()=> {}); //true
isFunction(()=> {return 1}); //true
isFunction(async function asyncFunction(){}); //true
isFunction(Array); //true
isFunction(Date); //true
isFunction(Object); //true
isFunction(Number); //true
isFunction(String); //true
isFunction(Symbol); //true
isFunction({}); //false
isFunction([]); //false
isFunction("function"); //false
isFunction(true); //false
isFunction(1); //false
isFunction("Alireza Dezfoolian"); //false

Answer 10

An other simply way:

var fn = function () {}
if (fn.constructor === Function) {
  // true
} else {
  // false
}

Answer 11

For those who's interested in functional style, or looks for more expressive approach to utilize in meta programming (such as type checking), it could be interesting to see Ramda library to accomplish such task.

Next code contains only pure and pointfree functions:

const R = require('ramda');

const isPrototypeEquals = R.pipe(Object.getPrototypeOf, R.equals);

const equalsSyncFunction = isPrototypeEquals(() => {});

const isSyncFunction = R.pipe(Object.getPrototypeOf, equalsSyncFunction);

As of ES2017, async functions are available, so we can check against them as well:

const equalsAsyncFunction = isPrototypeEquals(async () => {});

const isAsyncFunction = R.pipe(Object.getPrototypeOf, equalsAsyncFunction);

And then combine them together:

const isFunction = R.either(isSyncFunction, isAsyncFunction);

Of course, function should be protected against null and undefined values, so to make it "safe":

const safeIsFunction = R.unless(R.isNil, isFunction);

And, complete snippet to sum up:

const R = require('ramda');

const isPrototypeEquals = R.pipe(Object.getPrototypeOf, R.equals);

const equalsSyncFunction = isPrototypeEquals(() => {});
const equalsAsyncFunction = isPrototypeEquals(async () => {});

const isSyncFunction = R.pipe(Object.getPrototypeOf, equalsSyncFunction);
const isAsyncFunction = R.pipe(Object.getPrototypeOf, equalsAsyncFunction);

const isFunction = R.either(isSyncFunction, isAsyncFunction);

const safeIsFunction = R.unless(R.isNil, isFunction);

// ---

console.log(safeIsFunction( function () {} ));
console.log(safeIsFunction( () => {} ));
console.log(safeIsFunction( (async () => {}) ));
console.log(safeIsFunction( new class {} ));
console.log(safeIsFunction( {} ));
console.log(safeIsFunction( [] ));
console.log(safeIsFunction( 'a' ));
console.log(safeIsFunction( 1 ));
console.log(safeIsFunction( null ));
console.log(safeIsFunction( undefined ));

However, note the this solution could show less performance than other available options due to extensive usage of higher-order functions.

Answer 12

If you use Lodash you can do it with _.isFunction.

_.isFunction(function(){});
// => true

_.isFunction(/abc/);
// => false

_.isFunction(true);
// => false

_.isFunction(null);
// => false

This method returns true if value is a function, else false.

Answer 13

I found that when testing native browser functions in IE8, using toString, instanceof, and typeof did not work. Here is a method that works fine in IE8 (as far as I know):

function isFn(f){
    return !!(f && f.call && f.apply);
}
//Returns true in IE7/8
isFn(document.getElementById);

Alternatively, you can check for native functions using:

"getElementById" in document

Though, I have read somewhere that this will not always work in IE7 and below.

Answer 14

The below seems to work for me as well (tested from node.js):

var isFunction = function(o) {
     return Function.prototype.isPrototypeOf(o);
};

console.log(isFunction(function(){})); // true
console.log(isFunction({})); // false

Answer 15

I think you can just define a flag on the Function prototype and check if the instance you want to test inherited that

define a flag:

Function.prototype.isFunction = true; 

and then check if it exist

var foo = function(){};
foo.isFunction; // will return true

The downside is that another prototype can define the same flag and then it's worthless, but if you have full control over the included modules it is the easiest way

Answer 16

you should use typeOf operator in js.

var a=function(){
    alert("fun a");
}
alert(typeof a);// alerts "function"

Answer 17

Since node v0.11 you can use the standard util function :

var util = require('util');
util.isFunction('foo');

Answer 18

Becareful about this :

typeof Object === "function" // true.