Typescript string driven typings

Question

Is it possible to declare typing for a Typescript function with the following behavior:

declare const fun: HowToTypeThis


export const string_or_number_fun = fun<
  'num', number,
  'str', string
  >()

string_or_number_fun should accept either
string_or_number_fun('str', string)
or
string_or_number_fun('num', number)

string_or_number_fun('str', '') 
string_or_number_fun('num', 12)

My attempts led me to this:

interface HowToTypeThis {

  <Name1 extends string, Type1,
    Name2 extends string, Type2>(): <N = Name1 | Name2>(_: N, __: N extends Name1
      ? Type1
      : N extends Name2
      ? Type2
      : never): any

}

TS compiler reports string_or_number_fun as of type:

 const string_or_number_fun: 
 <N = "num" | "str">(_: N, __: N extends "num" ? number : N extends "str" ? string : never) => any

and it looks ok

but i get :
[ts] Argument of type '""' is not assignable to parameter of type 'never'. [2345]
and
[ts] Argument of type '12' is not assignable to parameter of type 'never'. [2345]

and it doesn't either guard from typing an unexpected string as first argument

casieber · Accepted Answer · 2018-11-30 22:29:28Z

2

You can achieve what you are looking for by first declaring a mapping of your custom type names to the type that it represents and then using that as part of your function declaration. Something like:

interface TypeNameToType {
    num: number;
    str: string;
}

function fun<T extends keyof TypeNameToType>(type: T, value: TypeNameToType[T])
{
    return value;
}

Now if you try and call fun('num', 'thisisastring'); Typescript will get mad that you used 'num' but the second param is not a number.

answered Nov 30, 2018 at 22:29

casieber

7,6122 gold badges23 silver badges35 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

aleclofabbro Over a year ago

nice one, this solution doesn't follow my construct, though, however, looks like the mapping interface thing could be introduced, to make it neater

artem · Accepted Answer · 2018-11-30 22:30:17Z

1

There must be a constraint for N type parameter, otherwise it will be widened to string, resulting in never coming from a conditional type.

N extends Name1 | Name2 is the only missing part in your code:

interface HowToTypeThis {

  <Name1 extends string, Type1,
    Name2 extends string, Type2>(): <N extends Name1 | Name2 = Name1 | Name2>(_: N, __: N extends Name1
      ? Type1
      : N extends Name2
      ? Type2
      : never) => any

}

answered Nov 30, 2018 at 22:30

artem

52.2k10 gold badges86 silver badges85 bronze badges

2 Comments

aleclofabbro Over a year ago

Wow, wonder why I didn't try that! :) 1 q: I now tried <N extends Name1 | Name2 > instead of <N extends Name1 | Name2 = Name1 | Name2> and it seems to be ok too, any consideration about this ?

artem Over a year ago

it seems that the the default value for N, = Name1 | Name2, is unnecessary here; normally its used when you want to have fallback value for the type parameter if it's not provided at the call site. Default is not needed if the compiler can infer correct type for N from argument values.

Collectives™ on Stack Overflow

Typescript string driven typings

2 Answers 2

1 Comment

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related