Typescript omit argument from generic function type-CodePudding

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({...})   // the generic is no longer valid, bit is should be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?

CodePudding user response：

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