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 consider objects for user data. 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, we'll get an error:

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

alert(user.address.street); // Error!

That's the expected result, JavaScript works like this, but many practical cases we'd prefer to get undefined instead of an error (meaning "no street").

...And another example. In the web development, we may need to get an information about an element on the page, that sometimes doesn't exist:

// Error if the result of querySelector(...) is null
let html = document.querySelector('.my-element').innerHTML;

Before ?. appeared in the language, the && operator was used to work around that.

For example:

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

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

AND'ing the whole path to the property ensures that all components exist (if not, the evaluation stops), but is cumbersome to write.

Optional chaining

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

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

Here's the safe way to access user.address.street:

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

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

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.

In the example above, user?. allows only user to be null/undefined.

On the other hand, if user does exist, then it must have user.address property, otherwise user?.address.street gives an error at the second dot.

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

For example, if according to our coding logic `user` object must be there, but `address` is optional, then `user.address?.street` would be better.

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;

There must be a declaration (e.g. let/const/var user). 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 user1 = {
  admin() {
    alert("I am admin");
  }
}

let user2 = {};

*!*
user1.admin?.(); // I am admin
user2.admin?.();
*/!*
```

Here, in both lines we first use the dot (`user1.admin`) to get `admin` property, because the user object must exist, so it's safe read from it.

Then `?.()` checks the left part: if the admin function exists, then it runs (that's so for `user1`). Otherwise (for `user2`) 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 user1 = {
  firstName: "John"
};

let user2 = null; // Imagine, we couldn't authorize the user

let key = "firstName";

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

alert( user1?.[key]?.something?.not?.existing); // 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:

obj?.prop -- returns obj.prop if obj exists, otherwise undefined.
obj?.[prop] -- returns obj[prop] if obj exists, otherwise undefined.
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 to exist. So that it won't hide programming errors from us, if they occur.

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Optional chaining '?.'

The "non-existing property" problem

Optional chaining

Summary

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