Convert to optional chain expression action creates incorrect code

[rubiesonthesky]

🔎 Search Terms

Convert to optional chain

🕗 Version & Regression Information

• This changed between versions ______ and ______
• This changed in commit or PR _______
• This is the behavior in every version I tried, and I reviewed the FAQ for entries about _________
• I was unable to test this on prior versions because playground doesn't support code actions

I'm able to repro this with versions 5.8.3 and 5.9.2 and VS Code Version: 1.104.3

⏯ Playground Link

https://www.typescriptlang.org/play/?#code/JYOwLgpgTgZghgYwgAgCJzHZBvAUM5AawgE91MB+ALmQGlTy4BuXAX111ElkRXrIxY8BYiRoBnMFFABzFgQAmEcQmkAHMMAD2IasknSQcthyUIANnCgoEOycgWCajFrZD3RjAMpTZyALwOggB0noLIFEGYoQwhosgA1MgA5FTJicgAFI7RYdFKKuqaOhGRyckAlMg05UwE9Q2NTc0tLbgA9O2t3T29jQB6FBydyABKygCu5mDIWjDIYAAWKIjFIDQdXW4esZg+hjIBUXAxAnu+RhSnpSnJTEA

💻 Code

interface Data {
  keyData?: KeyData;
}

interface KeyData {
  key: string;
  description?: string;
}

declare const data: Data;
const keyDataString = data.keyData ? data.keyData.key + ':' + (data.keyData.description ?? '') : '';

Try to invoke "Convert to optional chain expression" code action before colon.

🙁 Actual behavior

This is result of the action

const keyDataString = data.keyData?.key ?? '';

🙂 Expected behavior

It should not offer the action here as it's not a simple case.

Additional information about the issue

I was able to repro this with the tsconfig settings used in Playground but also in VS Code editor without tsconfig file.