Array that accepts the type of the first element as its generic parameter

There was a type challenge

The task was to create a generic function called First that takes an array T and returns the type of its first element.

type arr1 = ["a", "b", "c"];
type arr2 = [3, 2, 1];

type head1 = First<arr1>; // expected: 'a'
type head2 = First<arr2>; // expected: 3

I found it confusing as the requirement was to return the type of the first element. So shouldn't the output be string for the first case and number for the second?

What am I missing?

P.S. The given solution returns "a" instead of string.

type First<T extends any[]> = T extends [] ? never : T[0];

Update:

In the following code snippet, head1 is inferred as string.

let arr1 = ["a", "b", "c"]; // using let instead of type
type head1 = First<typeof arr1>; 

Why is it that when I use type array1 as in my initial question, head1 is inferred as "a", but when using above code, it's inferred as "string"?

Answer №1

The main point to understand is that arr1 and arr2 are not regular arrays; they are known as tuple types. In JavaScript, tuples are essentially arrays, but in TypeScript's type system, each element of a tuple has its own specific type.

This concept becomes clearer when you utilize "actual" types instead of simple literals, like this:

type arr1 = [string, number, Object];

const arr1Value: arr1 = ["hello", 42, {}];

type head1 = First<arr1>; // string

It's important to note that arr1 is not an ordinary array. It represents a type of array (or tuple), while arr1Value is an actual array belonging to that specific type.

If you do not explicitly provide a type for an array literal, TypeScript will infer it for you:

let arr1 = ["a", "b", "c"]; // arr1 is considered as type string[]

In some cases, the inferred type of arr1 could have been interpreted as ['a', 'b', 'c']; a tuple consisting of those three strings. However, this may restrict certain operations like pushing additional values onto the array. Therefore, TypeScript assumes it as a string[] by default. (If you specifically want to infer a tuple type, simply add as const at the end.)

Answer №2

In the world of TypeScript, type declarations are not limited to just string. While string is a valid type, it is considered quite broad.

To narrow down the possible values to only include 'a', 'b', or 'c', you can use the following syntax:

type Abc = 'a' | 'b' | 'c';

If you attempt to assign a value outside of these options to a variable declared with this type, TypeScript will flag it as an error during compilation:

const name: Abc = 'John'; // This will be marked as incorrect by TypeScript

A simple solution to your problem involves the following code snippet:

type First<T extends any[]> = T[0];

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