#`typeof` declines anonymous function
wind rivet
I thought that would be possible to give typeof an anonymous function
it seems, typescript does not work the way I wanted
I'd like to get an explanation why it's not possible. I would not like to create an individual function just to feed it to a typeof and never use
typeof declines anonymous function
proper chasm
typeof can only be used on an identifier, not an expression.
Instead of writing ReturnType<typeof () => db.fn()> you can do something like ReturnType<(typeof db)['fn']>.
quaint path
or just
const someFn = () => db.fn()
type FnReturnType = ReturnType<typeof someFn>
```since the function will not be called?yeah.. buuut it will hang in the compiled js, right? 👉👈
quaint path
yeah 😓
wind rivet
misty stump
Maybe you can provide more context about what you're actually trying to do?
wind rivet
Maybe you can provide more context about what you're actually trying to do?
I simply wanted to get the awaited type of a prisma query with complex nested selectors, though I was not going to use this function. Sounds weird but that's what it is
misty stump
@wind rivet Well, if you don't mind a dirty hack:
const hackery = (false as true) && db.fn(/*args*/);
type YourType = typeof hackery;
!resolved
karmic umbra
@wind rivet If you write out the function and hover over it, you can see the generic type arguments that have been inferred
Then you can just write typeof function with that generic type argument
no hacks required
quaint path
oh, was instantiation expression available on typeof?
karmic umbra
It's not an instantiation expression. It's just a generic type.
typeof func is a generic type, and we can use Generic<T> with that generic type, ReturnType<GenericFunction<T>> in this case.
quaint path
it seems like it is an instantiation expression since instantiation expressions are released on TS 4.7 🤔
quaint path
on TS 4.6, it is a syntax error
molten fogBOT
sandiford#0
karmic umbra
// TS 4.6.4
declare const f: <T>() => T
type F = typeof f // <T>() => T
type T = F<0> // Type 'F' is not generic.(2315)
type F<T> = <U>() => U & T
// ^? - type F<T> = <U>() => U & T
type F1 = F<0>
// ^? - type F1 = <U>() => U & 0