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Mutable and immutable useRef semantics with React & TypeScript

Wojciech Matuszewski
Doing stuff with JavaScript / Golang. Probably coding / working or at the gym.
・3 min read

In this post, you will learn how different ways declaring a ref with useRef hook influence the immutability of the current ref property. We will be looking at how to make the current property immutable, mutable, and know without much effort if the ref is one or the other.

All the behavior I'm going to talk about is only relevant in the context of TypeScript. The mutability / immutability is enforced at type level, not runtime level.

Immutable current property

The immutable semantics of the useRef hooks are usually used with DOM elements. A common use-case might be to get the ref of an element and focus that element whenever a button is clicked.

Here is how I would write that.

import * as React from "react";

const Component = () => {
  const inputRef = React.useRef<HTMLInputElement>(null);

  return (
    <div>
      <input type="text" name="name" ref={inputRef} />
      <button type="button" onClick={() => inputRef.current?.focus()}>
        Click to focus the input
      </button>
    </div>
  );
};
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Notice the type and the value I’ve initialized the useRef with. The semantics I’ve used signal that I’m relying on React to manage the ref for me. In our case, this means that I cannot mutate the inputRef.current. If I ever tried to do that, TypeScript would complain.

import * as React from "react";

const Component = () => {
  const inputRef = React.useRef<HTMLInputElement>(null);


  return (
    <div>
        {/* Cannot assign to 'current' because it is a read-only property */}
      <input type = "text" ref = {callbackRefValue => inputRef.current = callbackRefValue}>
      <button type="button" onClick={() => inputRef.current?.focus()}>
        Click to focus the input
      </button>
    </div>
  );
};
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After writing similar code for a while, I’ve created a rule of thumb I follow to understand if the ref that I’m looking is immutable.

If the useRef is initialized with null and the initial value does not belong to the provided type, the current property is immutable.

In our case, the null initial value does not belong to the type HTMLInputElement so the current property cannot be mutated.

Mutable current property

To have the current property of the ref be mutable, we need to change how we are declaring ref itself.

Suppose we are writing a component that deals with timers. The useRef hook is an ideal candidate to hold a reference to a timer. With the timer reference at hand, we can make sure that we clear the timer when the component unmounts.

Here is an, albeit a bit contrived, example.

import * as React from "react";

const Component = () => {
  const timerRef = React.useRef<number | null>(null);
  // This is also a valid declaration
  // const timerRef = React.useRef<number>()

    React.useEffect(() => {
        // Mutation of the `current` property
        timerRef.current = setTimeout(/* ... */)
        return clearInterval(timerRef.current)
    }, [])

  return (
      // ...
  );
};
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Since in the beginning, I have no way to know what the reference to the later declared setTimeout might be, I've initialized the useRef with null. Apart from the types, the declaration of the ref might seem eerily similar to the one in the Immutable current property section.
However, since the initially provided value (in our case null) wholly belongs to the type I've declared the useRef with (number | null), the current property is allowed to be mutable.

Similarly to the immutable current property case, here is my rule of thumb.

If the useRef is initialized with a value that belongs to the provided type, the current property of the ref is mutable.

In our case, the null initial value belongs to the type number | null so the current property can be mutated.
As an alternative, I could have declared the timerRef variable the following way

const timerRef = React.useRef<number>(); // the `timerRef.current` is also mutable
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Why is the current allowed to be mutated in this case? Because the timerRef is implicitly initialized with the undefined value. The undefined value belongs to the type I've declared the timerRef - the React.useRef typings are overloaded depending on the type of the initial value.

const timerRef = React.useRef<number>();

// Really is
const timerRef = React.useRef<number>(undefined);

// The `React.useRef` type definitions specify an overload whenever the type of the initial value is `undefined`
function useRef<T = undefined>(): MutableRefObject<T | undefined>; // Notice the `MutableRefObject`.
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Summary

When I started working with React & TypeScript, I found the difference between mutable and immutable refs quite confusing. I hope that this article was helpful and cleared some of the questions you might have had on the subject matter.

You can find me on twitter - @wm_matuszewski.

Thank you for your time.

Discussion (8)

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maciekgrzybek profile image
Maciek Grzybek

Nice 😊II like reading about these kinds of nuances 👌

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fchaplin profile image
Frederic CHAPLIN

Thank you ! A question : your constants are declared as any. Shouldn't we type every declaration in Typescript ?

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wojciechmatuszewski profile image
Wojciech Matuszewski Author

Hey, thank you for reaching out.

In TypeScript, you can leverage type inference. So, while I could explicitly annotate every variable with the corresponding type, I defer that work until necessary and rely on the notion of type inference.

You can read more about type inference here: typescriptlang.org/docs/handbook/t...

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fchaplin profile image
Frederic CHAPLIN • Edited

Yep, but it may be a better practice to explicitly type your constants while declaring them, especially when inferred type is not basic. You may gain time on weird issues later. And your code may be clearer when debugging 🙂.

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nicholasboll profile image
Nicholas Boll • Edited

I think explicitly typing everything makes it harder to read and now you have to understand the nuances of when you should or should not explicitly type.

const foo = 'foo' // 'foo'
const bar: string = 'bar' // string
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Not only is the second one harder to read and parse, it actually widened our type.

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fchaplin profile image
Frederic CHAPLIN • Edited

This is a really simple assignation example and I agree with you on this (except for a little typo) . But for function returns, and libs specific types, making a rule of typing explicitly everything WILL help.

Example:

const timerRef : React.MutableRefObject<number | null> = React.useRef<number | undefined>();
//mutable
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or

const inputRef: React.RefObject<HTMLInputElement> = React.useRef<HTMLInputElement>(null);
//not mutable
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By explicitly typing, you give explicitly the mutability information to other developpers (or to you in a month or two). So you improve readability.

And if you try

const inputRef: React.MutableRefObject<HTMLInputElement> = React.useRef<HTMLInputElement>(null);
//TS2322: Type 'RefObject<HTMLInputElement>' is not assignable to type MutableRefObject<HTMLInputElement>'.
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Here, typescript tell you instantly you're making a mistake: "No, it's not mutable!".

I know there are many sources that says you can use implicit types, but if you use them too much, you may lose some typescript gifts.

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nicholasboll profile image
Nicholas Boll • Edited

I'd probably argue there should be a useMutableRef and useRef rather than complicated types to communicate intent. I often have these small functions that map to normal functions to more clearly communicate intent:

const mutableRef = useMutableRef(false) // mutable, default assigned
const immutableRef = useRef<HTMLInputElement>(null) // React handles this, no default assigned

/**
 * Alias to `useEffect` that intentionally is only run on mount/unmount
 */
const useMount = (callback?: () => void) => {
  React.useEffect(callback, [])
}
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It is even possible to create nice utility functions that make element refs easier to work with:

// util file
function useElementRef<E extends keyof ElementTagNameMap>(element: E) {
  return React.useRef<ElementTagNameMap[E]>(null)
}

// usage
const ref = useElementRef('div') // React.RefObject<HTMLDivElement>
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Notice the theme where the Typescript types start to disappear for normal usage? This means you can still get the benefits of Typescript without explicitly using Typescript. Even JavaScript users of your code can benefit. This technique works better for libraries, especially if you have JavaScript users of your library. You can use JSDoc to explicitly type JS code, but that is a pain for non-primitive types.

I say there doesn't need to be a tradeoff between Typescript gifts and expressing intent. If your team only uses Typescript and understands all the types in use, maybe you don't need to spend any extra time communicating intent through functions. But it is very useful for JavaScript users in addition to Typescript users who don't spend time finding out all the nuances of Typescript type differences like useRef. You have to learn something extra either way (type differences or which function to use), but why not communicate intent explicitly through names vs types?

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kazamov profile image
kazamov

Thanks for the explanation!