Checking for passed parameters using null - JavaScript

Question

Take an example function here:

function a(b){
    console.log(b != null ? 1 : 2);
}

That code works fine, by printing 1 if you pass a parameter, and 2 if you don't.

However, JSLint gives me a warning, telling me to instead use strict equalities, i.e !==. Regardless of whether a parameter is passed or not, the function will print 1 when using !==.

So my question is, what is the best way to check whether a parameter has been passed? I do not want to use arguments.length, or in fact use the arguments object at all.

I tried using this:

function a(b){
    console.log(typeof(b) !== "undefined" ? 1 : 2);
}

^ that seemed to work, but is it the best method?

Answer 1

When no argument is passed, b is undefined, not null. So, the proper way to test for the existence of the argument b is this:

function a(b){
    console.log(b !== undefined ? 1 : 2);
}

!== is recommended because null and undefined can be coerced to be equal if you use == or !=, but using !== or === will not do type coercion so you can strictly tell if it's undefined or not.

Answer 2

You can use the falsy nature of undefined (a parameter, which was not passed is in fact undefined, not null) and just write:

(!b)?1:2

However this will also be true for 0, null and "" (falsy values).

If you want to write it the bulletproof way, you can go:

typeof(b) === "undefined"
// or thanks to the write protection for undefined in html5
b === undefined

Update: thanks to EcmaScript 2015, we now can use default parameters:

function a(b = 1){
    console.log(b);
}

If a parameter is undefined - either ommited or explicitely handed over (here you should use null instead) - the default value will be used, all other values remain unchanged (also falsy ones). Demonstration