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Don't use Function type in TypeScript

Matt Pocock
Matt PocockMatt is a well-regarded TypeScript expert known for his ability to demystify complex TypeScript concepts.

Quick Fix

  • You shouldn't use Function as a type. It represents any function.

  • Usually, you want to be more specific - like specifying the number of arguments, or what the function returns.

  • If you do want to represent a function that can take any number of arguments, and return any type, use (...args: any[]) => any.

Full Explanation

Let's imagine you're creating a function which sums up an array of objects. Here's one, taken from the Excalidraw codebase:

ts
const sum = <T>(
array: readonly T[],
mapper: (item: T) => number
): number =>
array.reduce(
(acc, item) => acc + mapper(item),
0
);

Let's look at the type definition. This function takes in:

  • An array of something: readonly T[]
  • A mapper function: (item: T) => number

and returns number.

In the body, it calls array.reduce(func, 0). This means the acc in the function begins as 0.

For each member of the array, it then adds acc and mapper(item) together. So, you end up with the sum of all of the members of the array.

What's the function declaration doing?

The mapper function is the key. Let's strip it out to take a look at it:

ts
type Mapper<T> = (item: T) => number;

Let's imagine a use case for this:

ts
interface YouTubeVideo {
name: string;
views: number;
}
 
const youTubeVideos: YouTubeVideo[] = [
{
name: "My favorite cheese",
views: 100,
},
{
name: "My second favorite cheese (you won't believe it)",
views: 67,
},
];
 
const mapper: Mapper<YouTubeVideo> = (video) => {
return video.views;
};
 
const result = sum(youTubeVideos, mapper); // 167

Here, mapper represents the function that extracts the number from the object. The powerful thing about the sum function is that you can discard most of these type declarations:

ts
const youTubeVideos = [
{ name: "My favorite cheese", views: 100 },
{
name: "My second favorite cheese (you won't believe it)",
views: 67,
},
];
 
const result = sum(youTubeVideos, (video) => {
return video.views;
}); // 167

We've actually discarded all of the type declarations, but video is still inferred as { name: string; views: number }. This is possible because of the specificity of our function definition: (item: T) => number.

What about Function?

The big mistake I see a lot of beginner devs making is declaring a function like mapper with the Function type:

ts
const sum = <T>(
array: readonly T[],
mapper: Function
): number =>
array.reduce(
(acc, item) => acc + mapper(item),
0
);

This keyword basically stands for 'any function'. It means that sum can technically receive any function.

When used in sum, we lose a lot of the safety that (item: T) => number provided:

ts
const result = sum(youTubeVideos, (item) => {
Parameter 'item' implicitly has an 'any' type.7006Parameter 'item' implicitly has an 'any' type.
// We can return anything from here, not just
// a number!
return item.name;
});

TypeScript now can't infer what item is supposed to be, or what our mapper function is supposed to return.

The lesson here is 'don't use Function' - there's always a more specific option available.

Alternatives

Expressing 'any function'

Sometimes, you'll want to express 'any function' in TypeScript. For this, let's look at some of TypeScript's built-in types, Parameters and ReturnType.

ts
export type Parameters<
T extends (...args: any) => any
> = T extends (...args: infer P) => any
? P
: never;
 
export type ReturnType<
T extends (...args: any) => any
> = T extends (...args: any) => infer R ? R : any;

You'll notice that both of these utility types use the same constraint: (...args: any) => any.

(...args: any) specifies that the function can take any number of arguments, and => any indicates that it can return anything.

Expressing a function with no args

For expressing a function with no arguments (but that returns anything), you'll want to use () => any:

ts
const wrapFuncWithNoArgs = (func: () => any) => {
try {
return func();
} catch (e) {}
};
 
wrapFuncWithNoArgs((a: string) => {});
Argument of type '(a: string) => void' is not assignable to parameter of type '() => any'. Target signature provides too few arguments. Expected 1 or more, but got 0.2345Argument of type '(a: string) => void' is not assignable to parameter of type '() => any'. Target signature provides too few arguments. Expected 1 or more, but got 0.

Summary

Function should never be used when expressing types.

(a: string, b: number) => any syntax can be used when you want to specify only the arguments, but not the return type.

(...args: any) => any can be used to represent any function type.

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