0014 · First of Array

Challenge Link

Problem

Implement a generic First<T> that takes an array T and returns its first element's type.

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

type head1 = First<arr1> // expected to be 'a'
type head2 = First<arr2> // expected to be 3

Solution

// Option 1: Conditional type approach
type First<T extends any[]> = T extends [] ? never : T[0]

// Option 2: Infer approach
type First<T extends readonly any[]> = T extends [infer F, ...infer R] ? F : never

Explanation

Naive Approach and Its Problem

The simplest idea is to just index the first element:

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

This works for non-empty arrays, but when T is an empty array [], T[0] resolves to undefined instead of never. The test cases expect never for empty arrays.

Option 1: Conditional Type Guard

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

We explicitly check if T is an empty array. If it is, return never; otherwise return T[0].

Step by step:

1. T extends any[] — constrains T to be an array type
2. T extends [] — checks if T is the empty tuple type
3. If empty → never; otherwise → T[0] (the first element)

Option 2: The infer Keyword

type First<T extends readonly any[]> = T extends [infer F, ...infer R] ? F : never

This uses the infer keyword to pattern-match the tuple structure:

• [infer F, ...infer R] matches any non-empty tuple, binding the first element to F and the rest to R
• If T matches (i.e., it's non-empty), return F
• If T is empty, the pattern doesn't match → return never

The infer keyword introduces a new type variable within a conditional type, letting TypeScript infer it from the matched structure.

Tips

readonly any[] is used instead of any[] so that the type works for both regular arrays and readonly tuples. A regular array is a subtype of a readonly array, so readonly any[] is the more general constraint.

Key concepts:

• Conditional Types — T extends U ? A : B
• Inferring Within Conditional Types — the infer keyword

中文解析

核心思路：

取数组第一个元素类型，看似简单，但需要处理空数组边界情况。

// 方案一：条件类型判断
type First<T extends any[]> = T extends [] ? never : T[0]

// 方案二：infer 模式匹配（更优雅）
type First<T extends readonly any[]> = T extends [infer F, ...infer R] ? F : never

为什么不能直接用 T[0]？

type First<T extends any[]> = T[0]  // ❌ 空数组时返回 undefined，而非 never

空元组 [] 的索引 [0] 类型是 undefined，但题目期望返回 never。

方案一解析：

• T extends [] 检查 T 是否为空元组
• 如果是空元组 → 返回 never
• 否则 → 返回 T[0]（第一个元素）

方案二解析（推荐）：

• [infer F, ...infer R] 是元组模式匹配
• infer F 绑定第一个元素的类型
• ...infer R 绑定剩余元素的类型（这里用不到）
• 空元组不匹配此模式 → 返回 never

考察知识点：

• 条件类型（Conditional Types）
• infer 关键字（类型推断）
• 元组模式匹配（Tuple Pattern Matching）