With two (or more) optional type parameters, when the second type argument depends on the first, and the function is called with one type argument, the second is improperly inferred.

[alex-eliot]

🔎 Search Terms

first type argument second type argument optional cannot be inferred
optional type argument wrongly inferred
when first type argument is passed, second one is not inferred

🕗 Version & Regression Information

Occurs on 5.4.0

⏯ Playground Link

https://www.typescriptlang.org/play?ts=5.4.0-dev.20240113#code/GYVwdgxgLglg9mABBATgUwIZTQeQEYBWa0APAFCKIAqiaAHtmACYDOiLUKMYA5ogLztO3HgBoKiHLQZpmbAErE4KJiSqjEACgCUAgHyIMYAJ4HBiiMtXqtu-gaOmyBzQGs0xlgC5qAbQC6GnCEPji6AN4S6FAgKEjBBGQAvmRklmAciOhsgqiY2PhE0Jq+AOTAcHClGqV4GCilgYiRlBVwPjr6iKUYLEzVEnUoHXYGPX2lydqpAPQzyAiZ2T6Rc5SIbSNdHFy8ANxka5RDW-ZCuzwHaylpi1BZaCwATALI6Fi4hMRQJOWVpYgAD7dIalPQlP5VGqgpotDaVU5jXr9cTHeqI7rIyZJaaHebpJaPJ4+CxWEgAIja5KBiHJQ3JGk6Z0ceiAA

💻 Code

function createObject<
 T extends string = string,
 O extends Record<T, () => any> = Record<T, () => any>
> (keys: T[], obj: O) {
 return obj
}

const res = createObject(['foo', 'bar'], {
 foo: () => 'asd',
 bar: () => 'asd'
})

// const res: {
// foo: () => string;
// bar: () => string;
// }

const res2 = createObject<'foo' | 'bar'>(['foo', 'bar'], {
 foo: () => 'asd',
 bar: () => 'asd'
})

// const res2: Record<"foo" | "bar", () => any>

🙁 Actual behavior

When the first type argument is passed, the second one is not inferred properly.

🙂 Expected behavior

The inferrence of the two code examples should be identical.

Additional information about the issue

No response

[MartinJohns]

This is working as intended. Type argument inference is all or nothing. You want #26349.

[typescript-bot]

This issue has been marked as "Duplicate" and has seen no recent activity. It has been automatically closed for house-keeping purposes.