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Optional Chaining

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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); // Error!

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 the 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 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. For more deeply nested properties, that becomes a problem as more repetitions are required.

E.g. let's try getting user.address.street.name.

We need to check both user.address and user.address.street:

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.

Don't even care to, as there's a 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.

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 ?..

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

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

So, if `user` happens to be undefined due to a mistake, we'll see a programming error about it and fix it. Otherwise, coding errors can be silenced where not appropriate, and become more difficult to debug.

````warn header="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 side effects, they don't occur.

For instance:

```js run
let user = null;
let x = 0;

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

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

let userGuest = {};

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

*!*
userGuest.admin?.(); // nothing (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.

```js run
let key = "firstName";

let user1 = {
  firstName: "John"
};

let user2 = null; 

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

Also we can use `?.` with `delete`:

```js run
delete user?.name; // delete user.name if user exists
```

````warn header="We can use `?.` for safe reading and deleting, but not writing"
The optional chaining `?.` has no use at the left side of an assignment.

For example:
```js run
let user = null;

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

It's just not that smart.

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 that the left part doesn't exist. So that it won't hide programming errors from us, if they occur.