How to initialize array to tuple type in TypeScript?

Question

I'm making sudoku web app with React/Redux. But I have encountered some problem with typings.

Current code:

// typedef
type Tuple9<T> = [T, T, T, T, T, T, T, T, T];

export type Board = Tuple9<Tuple9<number>>;

// code using board type, I want to fix getEmptyBoard() to more programmatic.
const getEmptyBoard: (() => Board) = () => {
  return [
    [0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0]
  ];
};

I want to fix getEmptyBoard() to more programmatic.

Is there good solution for this situation?
If 1, What is solution?

Shanon Jackson · Accepted Answer · 2019-06-04 00:39:18Z

1

Types don't exist at runtime meaning that there is no way to say "given type A build run-time Array X"

The more programmatic way to build it would be something like

new Array(9).fill(0).map(() => new Array(9).fill(0))

answered Jun 4, 2019 at 0:39

Shanon Jackson

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1 Comment

남현욱 Over a year ago

It is one of my solution but doesnt work cuz incompatible type of two.

ChrisF · Accepted Answer · 2025-02-05 18:43:56Z

0

For 9 I would do what you have done.

Otherwise you would follow the old functional programming saying : If its pure on the outside, it doesn't matter if its impure on the inside and make a strategic use of Tuple9 type assertion:

type Tuple9<T> = [T, T, T, T, T, T, T, T, T];
function make9<T>(what: T): Tuple9<T> {
    return new Array(9).fill(what) as Tuple9<T>;
}

export type Board = Tuple9<Tuple9<number>>;
function makeBoard(): Board {
    return make9(make9(0));
}

edited Feb 5 at 18:43

ChrisF

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answered Jun 4, 2019 at 0:40

basarat

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1 Comment

남현욱 Over a year ago

Thx :) const getEmptyBoard: (() => Board) = () => new Array(9).fill(0).map(() => new Array(9).fill(0)) as Board; This code works!

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