Type not narrowed when putting optional1 || optional2 condition in if statement and repeating it in its body

[verhovsky]

Suggestion

🔍 Search Terms

type inference in repeated expression
repeated or condition
or condition

✅ Viability Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript 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

This code (playground link):

function required(x: number): number {
  return x;
}
function optional(a?: number, b?: number): number {
  if (a || b) {
    return required(a || b);
  }
  return 0;
}

produces this type error on the return required(a || b) line:

Argument of type 'number | undefined' is not assignable to parameter of type 'number'.

but TypeScript should be able to infer that a || b can only be a number not number | undefined because the || expression must be true in the if block.

📃 Motivating Example

I am checking two optional properties of an object and using them interchangeably. Currently I have to use an as expression.

💻 Use Cases

See above.

[RyanCavanaugh]

Duplicate #35260