infer type assertion instead of boolean from "typeof" comparisons

[danielrentz]

Suggestion

🔍 Search Terms

infer type assertion typeof comparison

✅ Viability Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
  I am not sure abot this point, it may break some code...
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, new syntax sugar for JS, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

⭐ Suggestion

• The type of expressions typeof x === "<type>" will be inferred as x is <type> instead of boolean (x is Function for "function"; maybe x is object | null for "object"?).
• Type of expressions x instanceof Class will be inferred as x is Class instead of boolean.

📃 Motivating Example

Doing this would change the return type of custom type assertion functions, which in turn would help filtering arrays for specific type.

before:

const arr1: any[] = [];
const arr2 = arr1.filter(x => typeof x === "string"); // any[]

after:

const arr1: any[] = [];
const arr2 = arr1.filter(x => typeof x === "string"); // string[]

💻 Use Cases

Workaround is adding the type assertion manually:

const arr1: any[] = [];
const arr2 = arr1.filter((x): x is string => typeof x === "string"); // string[]

[MartinJohns]

Duplicate of #38390.

[danielrentz]

Thanks, must have slipped while searching with different terms through multiple pages of matches.