Typescript omit argument from generic function type

Question

How to omit prop type from generic function?

declare function func1<T extends {...}>(params: {age: number, something: T, ...more}): string

declare function func2<FunctionType extends (...args: any) => any>(cb: FunctionType): FunctionType
const newFunction = func2(func1)  // returns same type as func1
newFunction<{...}>({something: {...}, age: ...}) // is valid, but age should not be valid

in func2 how to remove property age from first argument of FunctionType?

I tried:

declare function func2<FunctionType extends (...args: any) => any>(cb: FunctionType): 
(params: Omit<Parameters<FunctionType>[0], 'age'>) => ReturnType<FunctionType>

const newFunction = func2(func1)  // This returns the new function type without age, but also without generics
newFunction<{...}>({...})   // the generic is no longer valid, bit is should be

This way I can remove the property age, but the return type is no longer generic. How I can keep it generic and omit the age from the first argument?

geoffrey · Accepted Answer · 2023-01-04 13:39:44Z

2

Not sure what you need that generic for since you don't seem to use it, but this should do:

declare function func2<P extends {}, R>(f: (params: P) => R)
   : (params: Omit<P, 'age'>) => R

answered Jan 4, 2023 at 13:39

geoffrey

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4 Comments

Oki Over a year ago

It is just an example, I have more generics in my real situation... anyway... I can't understand how this works. Your return type is (params: Omit<P, 'age'>) => R, and there are no generics in that type, but actually it accepts generics. How?

geoffrey Over a year ago

ReturnType and Parameters have the limitations you identified regarding generics, so instead of inferring the entire callback, which has generics, and loosing those generics, I inferred the parameter params and the return type as P and R directly. And this, TS can wire correctly.

Oki Over a year ago

I have to be able to add property to the returned function (just like an object) like this: const cb = () => {...}; cb.something = ... But this does not work. Do you have any solution for this (except cb.prototype.something = ...)? Since I return (params: Omit<P, 'age'>) => R, it does not allow me to add properties to the function

geoffrey Over a year ago

Theoretically you could return ((params: Omit<P, "age">) => R) & { foo: bar } but it doesn't seem to work unless you set the constraint of P to be {age: number, something: unknown, ...more} to match func1. I don't know why TS is restricting the variance as soon as we add that intersection, especially given it should only affect the return type. It would make for a good SO question.

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