31 replies

Use function as a parameter and preserve types

Hey guys, i'm trying to figure out a way to do the following

I have and interface with some functions/promises

interface SomeInterface {
  foo(...args): Promise<void>,
  bar(...args): void
  ...
}

interface SomeInterface {
  foo(...args): Promise<void>,
  bar(...args): void
  ...
}

Is there a way the create a function wrapper with the parameter is a function of

SomeInterface

SomeInterface

and the return type is the parameter of this function. Something like this:

function wrapper(func: FuncOfSomeInterface): ResultOfFunc {
  ...do smth here

  const result = func(...args)

  ...do smth here
  
  return result
}

function wrapper(func: FuncOfSomeInterface): ResultOfFunc {
  ...do smth here

  const result = func(...args)

  ...do smth here
  
  return result
}

Then i would do this:

wrapper(foo(...args))

wrapper(wrong(...args))

I still a noob in typescript, so i'd be happy if someone help me