Array.prototype.find not typing input as possibly undefined

[ChrisMcD1]

Bug Report

🔎 Search Terms

Array, find, holes, sparse array

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about Array.prototype.find

⏯ Playground Link

Playground link with relevant code

💻 Code

type Foo = {
  bar: number;
};

let arr: Array<Foo> = [];

arr[3] = {
  bar: 1,
};

const foundValue = arr.find((e) => e.bar === 1);

🙁 Actual behavior

TypeScript doesn't indicate that find has unique functionality among the Array.prototype methods in that it does not skip holes in an array. This seems to be intended behavior from what I have seen.

🙂 Expected behavior

TypeScript should mark that e could be of type Foo or undefined.

It doesn't make sense in my opinion to type arr as an Array<Foo | undefined> if it is known that you will never explicitly insert undefined into the array. Other javascript iterator functions like forEach will only go through the single defined element of arr, so the line arr.forEach((e) => e.bar ===1) would not throw a runtime exception. However, typing the array as Array<Foo | undefined> would result in a TypeScript error on that forEach statement.

[RyanCavanaugh]

TypeScript doesn't acknowledge the existence of sparse arrays; this particular function is only one of many which can get you into trouble. e.g. given the arr above, you could write for (const e of [...arr]) { e; } and have the same problem.

[ChrisMcD1]

Oh, okay! Thank you for informing me.