Get the type of the call signature from a literal object

[cangSDARM]

Bug Report

🔎 Search Terms

🕗 Version & Regression Information

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

⏯ Playground Link

Playground link with relevant code

💻 Code

// see playground

🙁 Actual behavior

I cannot get the type of its call signature

🙂 Expected behavior

Get the type of the call signatures from literal object

[MartinJohns]

You can just use: Omit<typeof pRetry, keyof typeof pRetry>

I'd argue this is working as intended. typeof pRetry includes the properties AbortError and default, which your assigned function does not provide.

[cangSDARM]

it works, but this doesn't look very crisp, and it's a very odd workaround if you ask me.

For other types that are slightly different, it doesn't seem to work

[cangSDARM]

there is a another solution, yet it's not support for generics:

type InferCall2<T> = T extends {
	(...args: infer Params): infer Ret 
} ? (...args: Params) => Ret : never

type pRetryCall =
	InferCall2<typeof pRetry>	// generics are gone

[typescript-bot]

This issue has been marked as 'Question' and has seen no recent activity. It has been automatically closed for house-keeping purposes. If you're still waiting on a response, questions are usually better suited to stackoverflow or the TypeScript Discord community.