为什么在只能返回窄型的映射之后，TypeScript不缩小型？

由 iut发布于 2020-05-18 16:14:06

In the code below I would expect the find function to return undefined | BufferEncoding, the same type as the Map has. But undefined | string is returned. How come?

  const chartdetToFsEncodings = new Map<string, BufferEncoding>([
    ["UTF-8", "utf8"],
    ["UTF-16LE", "utf16le"],
  ]);

  const supportedEncoding = analyse(buffer)
    .map((match) => match.name)
    .find((name) =>
      chartdetToFsEncodings.get(name)
    ) as BufferEncoding;

我尝试将Map设置为const，但是出现了一些语法错误。

  const chartdetToFsEncodings = new Map<string, BufferEncoding>([
    ["UTF-8", "utf8"],
    ["UTF-16LE", "utf16le"],
  ]) as const;

bmagni

2020-05-18 16:14:06

find method is used by an string[], so it will return string | undefined. If you want to have BufferEncoding | undefined, this is the easiest way:

const supportedEncoding = analyse(buffer)
    .map((match) => match.name)
    .find((name): name is BufferEncoding =>
      chartdetToFsEncodings.get(name) !== undefined
    );

零零柒

2020-05-18 16:14:06

The issue boils down to the incorrect use of Array.prototype.find. The method acts like a filter, but returns only the first element in the array for which the predicate returns true. In that case, it can either return an element from the mapped array, from .map(match => match.name), which is typed as string, or undefined.

If you want to get an array of BufferEncoding instead, you can do it directly in .map() callback:

const supportedEncoding = analyse(buffer)
    .map((match) => chartdetToFsEncodings.get(match.name))

iut

这家伙很懒，什么都没留下

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