147-cPrintfParser

题目描述

There is a function in C language: printf. This function allows us to print something with formatting. Like this:

printf("The result is %d.", 42);

This challenge requires you to parse the input string and extract the format placeholders like %d and %f. For example, if the input string is "The result is %d.", the parsed result is a tuple ['dec'].

Here is the mapping:

type ControlsMap = {
  c: 'char';
  s: 'string';
  d: 'dec';
  o: 'oct';
  h: 'hex';
  f: 'float';
  p: 'pointer';
};

分析

可以先看下用例：

type cases = [
  Expect<Equal<ParsePrintFormat<''>, []>>,
  Expect<Equal<ParsePrintFormat<'Any string.'>, []>>,
  Expect<Equal<ParsePrintFormat<'The result is %d.'>, ['dec']>>,
  Expect<Equal<ParsePrintFormat<'The result is %%d.'>, []>>,
  Expect<Equal<ParsePrintFormat<'The result is %%%d.'>, ['dec']>>,
  Expect<Equal<ParsePrintFormat<'The result is %f.'>, ['float']>>,
  Expect<Equal<ParsePrintFormat<'The result is %h.'>, ['hex']>>,
  Expect<Equal<ParsePrintFormat<'The result is %q.'>, []>>,
  Expect<Equal<ParsePrintFormat<'Hello %s: score is %d.'>, ['string', 'dec']>>,
  Expect<Equal<ParsePrintFormat<'The result is %'>, []>>,
];

其本质是匹配 %x，其中 x ，必须是 ControlsMap 中的属性。对于 %%x，可以判定为无效。

可以通过匹配 ${infer F}%${infer X}${infer R} 来进行处理。也可以遍历处理。

这里仅讲解匹配处理。

题解

type ParsePrintFormat<T extends string> =
  // 匹配 % 后的 X
  T extends `${infer F}%${infer X}${infer R}`
    ? // X 合法
      X extends keyof ControlsMap
      ? // 向结果中增加 X 对应的类型，并递归剩余字符
        [ControlsMap[X], ...ParsePrintFormat<R>]
      : // 不合法，直接递归剩余字符
        ParsePrintFormat<R>
    : // 没有匹配到则返回 []
      [];

知识点

1. 字符匹配套路