Type inference bails when encountering any in conditional expression

[mhofman]

Bug Report

🔎 Search Terms

type inference any conditional expression

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries

⏯ Playground Link

Playground link with relevant code

💻 Code

declare function getMaybeNumber(): any;
// declare function getMaybeNumber(): number | undefined;

function checkNum(num?: number) {
    if (!num) {
      num = getMaybeNumber() || 42;
    //   if (!num) num = 42
    }
    if (num < 50){
        throw new Error()
    };
}

🙁 Actual behavior

'num' is possibly 'undefined'.

🙂 Expected behavior

type of num narrowed to number or widened to any

More information

The narrowing work correctly if the return type is not any.

The narrowing also works if using an if statement instead of a conditional expressions. Likely related to the design limitation that was mentioned in #50739. However given that type narrowing works if the return type is not any there is obviously some type inference possible in conditional expressions.

[RyanCavanaugh]

The current behavior is the most-conservative since there's really no telling what is happening on that line of code, while still not completely wrecking type safety on num throughout the body of the function. I don't think a change in either direction as proposed is a net improvement.

[github-actions]

This issue has been marked as 'Not a Defect' and has seen no recent activity. It has been automatically closed for house-keeping purposes.