如何在 TypeScript 中配置更严格的类型检查？

Question

I want to have stricter code checks and I don't want to be allowed to have any implicit 'any' types. Therefore I enabled "strict": true

tsconfig.json

{
  "compilerOptions": {
    "target": "ES5",
    "module": "commonjs",
    "strict": true
  },
  "include": ["./"]
}

If I write this function:

const testing = () => console.log("test");

VSCode auto infers the returning type by it's usage:

const testing: () => void

Is there any way I can make this stricter and not allow VSCode to auto infer?

thanks

Answer 1

emmm, in this case, you'd ll better use eslint intead of hacking ts, see here:

https://github.com/typescript-eslint/typescript-eslint/blob/master/packages/eslint-plugin/docs/rules/explicit-function-return-type.md