Why does TypeScript behave this way? const isArray = (value: any)

Question

Why does TypeScript behave this way? const isArray = (value: any)

=>
Array.isArray(value)

const f = (
input: string | string[]
) => {
if (isArray(input)) {
// ✗ input: string | string[]
}
}

but

const g = (
input: string | string[]
) => {
if (Array.isArray(input)) {
// ✓ input: string[]
}
}

0

01.11.2019

1 ответов

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ᶜᵘᵇ Joshua ⱽᴵᴵ ⚔️🏹 𝖛𝖆𝖓 𝕵𝖆𝖓𝖉𝖆 · Accepted Answer

ᶜᵘᵇ Joshua ⱽᴵᴵ ⚔️🏹 𝖛𝖆𝖓 𝕵𝖆𝖓𝖉𝖆

Because your isArray is just a regular boolean function, so in f, TS doesnt have any hint about the type of input. On the other hand, Array.isArray seems to be a type guard, so if it evaluates to true, TS knows input is of type string[]. If you want your isArray to be equivalent to Array.isArray, define it as a custom type guard: const isArray(value: any): value is Array => Array.isArray(value) (im not sure if thats the exact syntax)

0

01.11.2019

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Why does TypeScript behave this way? const isArray = (value: any)

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