Question

我正在使用React / Redux制作数独Web应用程序。但是我在键入时遇到了一些问题。

当前代码：

// 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]
  ];
};

我想将getEmptyBoard()修改为更具编程性。

这种情况有好的解决方法吗？
如果为1，什么是解决方案？

Answer 1

类型在运行时不存在，这意味着无法说“给定类型A构建运行时数组X”

更编程的方式构建它就像

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

Answer 2

对于9，我会做你所做的。

否则，您将遵循古老的函数式编程：If its pure on the outside, it doesn't matter if its impure on the inside，并战略性地使用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));
}

如何在TypeScript中将数组初始化为元组类型？

2 个答案: