This Crazy Syntax Lets You Get An Array Element's Type
Learn how to extract the type of an array element in TypeScript using the powerful Array[number]
trick.
Let's look at a question I got asked the other day:
type TakesOneArgument = (a: string) => void
// Why does this not error in TypeScript?
const func: TakesOneArgument = () => {}
In the type TakesOneArgument
, we're saying that the function takes a string
and returns void
. But when we use TakesOneArgument
to annotate a function that doesn't take any arguments, TypeScript doesn't complain.
Why is this?
Let's look at an example that might look less weird:
const array = ['a', 'b', 'c']
// No errors!
array.forEach(() => {})
Here, we're using a .forEach
method on an array of strings. The function we pass to .forEach
could receive several arguments:
array.forEach((item, index, array) => {
// ...
})
But we're not using any of those arguments. So why doesn't TypeScript complain?
Let's imagine we've ditched TypeScript, and we're living in JavaScript-land again. We've got a function that takes a single argument:
function takesOneArg(a) {}
What's going to happen if we call it with two arguments?
takesOneArg('a', 'b')
Well - nothing! The second argument passed to it will be silently ignored, and it's unlikely ever to cause a runtime error (unless you're doing something unsafe with arguments
).
This is exactly what's happening in the .forEach
case:
The function passed to .forEach
is always passed item
, index
and array
- but it doesn't always need to specify them.
// Both are fine!
array.forEach((item, index, array) => {})
array.forEach(() => {})
So - when you specify a function type, TypeScript doesn't force you to handle all the parameters. It's perfectly fine to use a function that takes fewer arguments than the specified type.
Let's circle back to our first example:
type TakesOneArgument = (a: string) => void
// Why does this not error in TypeScript?
const func: TakesOneArgument = () => {}
The reason this doesn't error is because, technically, func
is assignable to TakesOneArgument
. You could use that function anywhere that TakesOneArgument
is expected, and it would work.
But as you can see, it's a little unexpected. The cleanest way to annotate this would be:
const func = (a: string): void => {}
I don't want to dissuade you from typing your functions using type
. But this behaviour is definitely worth considering.
Plus, I go into depth elsewhere on why function types can give you worse errors.
Function types are weird in TypeScript
Learn how to extract the type of an array element in TypeScript using the powerful Array[number]
trick.
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