打字稿忽略 null 或 undefined 作为函数返回的可能性

Question

I have a function where should return a LoginResponse type or null but Typescript lint is ignoring the null possibility and just displaying LoginResponse as possible value.

export const getSession = (): LoginResponse | null => {
  return null
}

I also tried using undefined without success. Any thoughts?

Answer 1

TLDR;

tsconfig.json:

{
  "compilerOptions": {
    "strict": true
  }
}

command line: tsc --strict

By default, TypeScript considers null and undefined to be subtypes of all types other types. That is to say, everything is nullable and or may not be defined.

Given two types, Base and Sub where Sub extends Base , their union, Base | Sub Base | Sub , is equivalent to just Base . Hence, since null is a subtype of all types, it is a subtype of LoginResponse making LoginResponse | null LoginResponse | null equivalent to LoginResponse

To have TypeScript treat null and undefined as distinct types, unrelated to all others you need to enable --strictNullChecks , a powerful TypeChecking option.

tsconfig.json

{
  "compilerOptions": {
    "strictNullChecks": true
  }
}

Or, on the command line, tsc --strictNullChecks.

The --strictNullChecks option is part of the strict family of powerful typechecking options and we can therefore use --strict , enabling all of them which is highly recommended, not just because it catches more errors and provides more semantics for those errors, but because it improves the quality of type inferences dramatically, making a lot of code that should intuitively work, in fact work.