How to apply a pre-defined type to a function in TypeScript
Published
Given the following type signature for a TypeScript function:
type BirthdayGreeter = (name: string, age: number) => string;
We have two primary options to apply this type signature to our function.
One is to use the type on a variable which is then assigned to an arrow function:
const greeter: BirthdayGreeter = (name, age) =>
`Congrats ${name}, you're ${age} years young!`;
The second is to use the type utilities Parameters
and ReturnType
to apply the types to the arguments (Parameters
) of the function and the return value.
function greeter(
...[name, age]: Parameters<BirthdayGreeter>
): ReturnType<BirthdayGreeter> {
return `Congrats ${name}, you're ${age} years young!`;
}
Parameters
grabs the arguments of the function type signature and returns an array of the types of each parameter, thus our need to use the spread (...
) operator to apply the parameters and then and then use the destructuring ([name, age]
) syntax to grab those arguments so we can then use them in the function body.
We also see ReturnType
which, as its name implies, grabs the type of a function type declaration (in this case string
) so we can then apply it to our function implementation return type.
Now, obviously the second example of trying to type a function
is much more verbose and has some confusing syntax.
If you’re trying to conform your function to an external contract (say a functional component in React or a Handle method in Svelte), using the arrow function syntax is much cleaner and likely preferable.
Heres to hoping that TypeScript develops a new way to apply types
For more details, checkout this great post by Dr. Axel Rauschmayer which goes much deeper into typing functions in TypeScript.
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