Optional chaining (?.)

Optional chaining (?.)

Optional chaining operator (?.) permits reading the value of a property located deep within a chain of connected objects without having to expressly validate that each reference in the chain is valid. The ?. operator functions similarly to the . chaining operator, except that instead of causing an error if a reference is nullish (null or undefined), the expression short-circuits with a return value of undefined. When used with function calls, it returns undefined if the given function does not exist.

The Problem 🤷🏻‍♀️

An object can have a very different nested structure of objects.

  • Fetching remote JSON data
  • Using configuration objects
  • Having optional properties

Working with data in JavaScript frequently involves situations where you aren’t sure that something exists. For example, imagine getting a JSON response from a weather API.

  "data": {
    "temperature": {
      "current": 68,
      "high": 79,
      "low": 45
    "averageWindSpeed": 8

You can go through each level of the object to get the high temperature. ⛅️

The value of response.data, response.data.temperature are confirmed to be non-null before accessing the value of response.data.temperature.current. This prevents the error that would occur if you simply accessed response.data.temperature.current directly without testing response.data && response.data.temperature

const highTemperature = response.data && response.data.temperature && response.data.temperature.current;

With the optional chaining operator(?.) you don't have to explicitly test and short-circuit based on the state of response.data && response.data.temperature before trying to access response.data.temperature.current.

If response.data && response.data.temperature are null or undefined, the expression automatically short-circuits, returning undefined.

const highTemperature = response.data?.temperature?.current;

By using the ?. operator instead of just ., JavaScript knows to implicitly check to be sure response.data && response.data.temperature are not null or undefined before attempting to access response.data.temperature.current

Optional chaining not valid on the left-hand side of an assignment

let object = {};
object?.property = 1; // Uncaught SyntaxError: Invalid left-hand side in assignment

Relationship with the optional chaining operator (?.)

The nullish coalescing operator treats undefined and null as specific values and so does the optional chaining operator (?.) which is useful to access a property of an object which may be null or undefined.

let foo = { someFooProp: "hi" };

console.log(foo.someFooProp?.toUpperCase());  // "HI"
console.log(foo.someBarProp?.toUpperCase()); // undefined

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

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

let user1 = {
  admin() {
    alert("I am admin");

let user2 = {};
user1.admin?.(); // I am admin

Use ?. with delete ␡:

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

Few Scenario which needs to taken care of:

  1. 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

There must be let/const/var user. The optional chaining works only for declared variables.

  1. Use ?. for safe reading and deleting, but not writing

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

// the idea of the code below is to write user.name, if user exists
user?.name = "John"; // Error, doesn't work
// because it evaluates to undefined = "John"

Summary 📝

The ?. 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 exists, otherwise returns undefined.

Reference 🧐

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