Type predicate for an optional property is not inferred

🔎 Search Terms

inferred type predicate optional

🕗 Version & Regression Information

• This changed between versions 5.4 and 5.5

⏯ Playground Link

https://www.typescriptlang.org/play/?ts=5.5.4#code/C4TwDgpgBA8gRgKygXigbwPZmASwwOwEMAbAfgC51DL8BXAWzggCcBfVgKFElkQCYU6LLgIkKUWvgAmEAGY58EKaygAfIdjxFilNNSh1GLdh1OzJAYxH4oOAM4xNo4gBE5CpQAoIleAgCU6BwA9MFQ4RFQAHqkHJHMEMC0zDYQAHTCWiRQAITIqJIy8opSHJwc5vhWWrYOTtpuxUp83r78gWghYTFxEQlJKVDpmc65+RLS7iVlHEA

💻 Code

type Obj = {optional?: {a: number}}

function isOptionalDefined(e: Obj) {
    return e.optional !== undefined
}

if (isOptionalDefined(someObj) {
  someObj.optional.a // 'someObj.optional' is possibly 'undefined'
}

🙁 Actual behavior

If a type contains an optional property, and I write a function that ensures the property is defined, the function is not inferred as a type predicate. No type narrowing happens when I use this function.

At the same time, if I rewrite the original type to be itself a union, then the type predicate is inferred, as expected.

🙂 Expected behavior

Type predicate is inferred

Additional information about the issue

I feel it has something to do with https://github.com/microsoft/TypeScript/issues/55257, but I'm not sure.

issue にリンクされている TypeScript Playground の再現コードから始め、optional-property predicate に対する 5.4 と 5.5 の動作を比較してください。関連するコンテキストについて issue #55257 の議論を読み、提供された例で predicate が推論され、optional property がエラーなしで絞り込まれれば完了です。