Optional chaining '?.'

A recent addition

This is a recent addition to the language. Old browsers may need polyfills.

The optional chaining ?. is a safe way to access nested object properties, even if an intermediate property doesn’t exist.

The “non-existing property” problem

If you’ve just started to read the tutorial and learn JavaScript, maybe the problem hasn’t touched you yet, but it’s quite common.

As an example, let’s say we have user objects that hold the information about our users.

Most of our users have addresses in user.address property, with the street user.address.street, but some did not provide them.

In such case, when we attempt to get user.address.street, and the user happens to be without an address, we get an error:

let user = {}; // a user without "address" property

alert(user.address.street);

That’s the expected result. JavaScript works like this. As user.address is undefined, an attempt to get user.address.street fails with an error.

In many practical cases we’d prefer to get undefined instead of an error here (meaning “no street”).

…and another example. In Web development, we can get an object that corresponds to a web page element using a special method call, such as document.querySelector('.elem'), and it returns null when there’s no such element.

// document.querySelector('.elem') is null if there's no element
let html = document.querySelector('.elem').innerHTML; // error if it's null

Once again, if the element doesn’t exist, we’ll get an error accessing .innerHTML property of null. And in some cases, when the absence of the element is normal, we’d like to avoid the error and just accept html = null as the result.

How can we do this?

The obvious solution would be to check the value using if or the conditional operator ?, before accessing its property, like this:

let user = {};

alert(user.address ? user.address.street : undefined);

It works, there’s no error… But it’s quite inelegant. As you can see, the "user.address" appears twice in the code.

Here’s how the same would look for document.querySelector:

let html = document.querySelector('.elem') ? document.querySelector('.elem').innerHTML : null;

We can see that the element search document.querySelector('.elem') is actually called twice here. Not good.

For more deeply nested properties, it becomes even uglier, as more repetitions are required.

E.g. let’s get user.address.street.name in a similar fashion.

let user = {}; // user has no address

alert(user.address ? user.address.street ? user.address.street.name : null : null);

That’s just awful, one may even have problems understanding such code.

There’s a little better way to write it, using the && operator:

let user = {}; // user has no address

alert( user.address && user.address.street && user.address.street.name ); // undefined (no error)

AND’ing the whole path to the property ensures that all components exist (if not, the evaluation stops), but also isn’t ideal.

As you can see, property names are still duplicated in the code. E.g. in the code above, user.address appears three times.

That’s why the optional chaining ?. was added to the language. To solve this problem once and for all!

Optional chaining

The optional chaining ?. stops the evaluation if the value before ?. is undefined or null and returns undefined.

Further in this article, for brevity, we’ll be saying that something “exists” if it’s not null and not undefined.

In other words, value?.prop:

• works as value.prop, if value exists,
• otherwise (when value is undefined/null) it returns undefined.

Here’s the safe way to access user.address.street using ?.:

let user = {}; // user has no address

alert( user?.address?.street ); // undefined (no error)

The code is short and clean, there’s no duplication at all.

Here’s an example with document.querySelector:

let html = document.querySelector('.elem')?.innerHTML; // will be undefined, if there's no element

Reading the address with user?.address works even if user object doesn’t exist:

let user = null;

alert( user?.address ); // undefined
alert( user?.address.street ); // undefined

Please note: the ?. syntax makes optional the value before it, but not any further.

E.g. in user?.address.street.name the ?. allows user to safely be null/undefined (and returns undefined in that case), but that’s only for user. Further properties are accessed in a regular way. If we want some of them to be optional, then we’ll need to replace more . with ?..

Don’t overuse the optional chaining

We should use ?. only where it’s ok that something doesn’t exist.

For example, if according to our code logic user object must exist, but address is optional, then we should write user.address?.street, but not user?.address?.street.

Then, if user happens to be undefined, we’ll see a programming error about it and fix it. Otherwise, if we overuse ?., coding errors can be silenced where not appropriate, and become more difficult to debug.

The variable before ?. must be declared

If there’s no variable user at all, then user?.anything triggers an error:

// ReferenceError: user is not defined
user?.address;

The variable must be declared (e.g. let/const/var user or as a function parameter). The optional chaining works only for declared variables.

Short-circuiting

As it was said before, the ?. immediately stops (“short-circuits”) the evaluation if the left part doesn’t exist.

So, if there are any further function calls or operations to the right of ?., they won’t be made.

For instance:

let user = null;
let x = 0;

user?.sayHi(x++); // no "user", so the execution doesn't reach sayHi call and x++

alert(x); // 0, value not incremented

Other variants: ?.(), ?.[]

The optional chaining ?. is not an operator, but a special syntax construct, that also works with functions and square brackets.

For example, ?.() is used to call a function that may not exist.

In the code below, some of our users have admin method, and some don’t:

let userAdmin = {
  admin() {
    alert("I am admin");
  }
};

let userGuest = {};

userAdmin.admin?.(); // I am admin

userGuest.admin?.(); // nothing happens (no such method)

Here, in both lines we first use the dot (userAdmin.admin) to get admin property, because we assume that the user object exists, so it’s safe read from it.

Then ?.() checks the left part: if the admin function exists, then it runs (that’s so for userAdmin). Otherwise (for userGuest) the evaluation stops without errors.

The ?.[] syntax also works, if we’d like to use brackets [] to access properties instead of dot .. Similar to previous cases, it allows to safely read a property from an object that may not exist.

let key = "firstName";

let user1 = {
  firstName: "John"
};

let user2 = null;

alert( user1?.[key] ); // John
alert( user2?.[key] ); // undefined

Also we can use ?. with delete:

delete user?.name; // delete user.name if user exists

We can use ?. for safe reading and deleting, but not writing

The optional chaining ?. has no use on the left side of an assignment.

For example:

let user = null;

user?.name = "John"; // Error, doesn't work
// because it evaluates to: undefined = "John"

Common mistakes

• Skipping the small examples and then missing the exact rule that Optional chaining '?.' depends on.
• Copying code without changing one value at a time to see which part controls the result.
• Treating similar-looking syntax or APIs as interchangeable before checking their edge cases.

Summary

The optional chaining ?. syntax has three forms:

1. obj?.prop – returns obj.prop if obj exists, otherwise undefined.
2. obj?.[prop] – returns obj[prop] if obj exists, otherwise undefined.
3. obj.method?.() – calls obj.method() if obj.method exists, otherwise returns undefined.

As we can see, all of them are straightforward and simple to use. The ?. checks the left part for null/undefined and allows the evaluation to proceed if it’s not so.

A chain of ?. allows to safely access nested properties.

Still, we should apply ?. carefully, only where it’s acceptable, according to our code logic, that the left part doesn’t exist. So that it won’t hide programming errors from us, if they occur.

Practice

• Rewrite one example from this lesson without looking at the original, then run it and compare the result.
• Change one input, operator, method call, or option in a code sample and predict what will happen before running it.
• Explain Optional chaining '?.' in your own words as if you were reviewing it with another learner.

Keep learning

Continue with Symbol type when you are ready for the next lesson.