打字稿中的递归表达式答案

【问题标题】：Recursive expression in typescript打字稿中的递归表达式
【发布时间】：2023-02-09 23:33:45
【问题描述】：

假设我想将一组逻辑规则存储在一个对象中：

const rule1 : Rule = {
  ">" : [3,2]
}; // Which represents : 3>2

const rule2 : Rule = {
  "and" : [
    { ">" : [3,1] },
    { "<" : [1,3] }
  ]
}; // Which represents : (3>1) && (1<3)

我这样写我的类型：

type Operand = number | string | Rule

type Operator =
  "var" |
  "===" |
  "!==" |
  "<="  |
  "<"   |
  ">="  |
  ">"   |
  "or"  |
  "and" ;

interface Rule extends Record<Operator, [Operand, Operand]> { }

但是我收到以下错误Type '{ ">": [number, number]; }' is missing the following properties from type 'Record<Operator, [Operand, Operand]>': var, "===", "!==", "<=", and 4 more.

我做错了什么？

【问题讨论】：

标签： typescript recursion types logic

【解决方案1】：

interface Rule extends Record<Operator, [Operand, Operand]> { } 将命名任何给定规则所需的所有运算符。

您可以扩展 Partial<Record<Operator, [Operand, Operand]>> 并将它们全部设为可选。这意味着一个规则可以指定多个运算符，这可能不是您想要的：

interface Rule extends Partial<Record<Operator, [Operand, Operand]>> { }
const rule1 : Rule = {
  ">" : [3, 2],
  "<" : [3, 2]
};

Playground Link

您也可以不使用接口，而是生成一个联合，其中每个对象类型都需要一个运算符：

type Rule<T extends Operator = Operator> =  T extends T ? Record<T, [Operand, Operand]>: never;

Playground Link

【讨论】：