How do I use compose in Typescript?

Question

Im having problems using compose in Typescript....

const rawHeaders = R.compose(
  R.join('\n'),
  R.map(x=>`${x[0]}: ${x[1]}`),
  R.toPairs
)

I have tried the below, but it's brutal. Anyone know a way to make this work more elegantly?

const rawHeaders:Function = R.compose(
  R.join('\n'),
  R.map((x:[String, String]) =>`${x[0]}: ${x[1]}`),
  (x:{s:String})=>R.toPairs(x))
)

I have also tried to use ts-ignore, which at present seems to be the best option.

const rawHeaders = R.compose(
  R.join('\n'),
  // @ts-ignore
  R.map(x=>`${x[0]}: ${x[1]}`),
  // @ts-ignore
  R.toPairs
)

Answer 1

Have you tried leveraging the typing of compose itself? You can give it the arguments and the return values of each function in the compose like so:

const rawHeaders = R.compose<
  { [key: string]: string | number }, // Argument
  Array<[string, string]>, // Return of toPairs
  string[], // Return of map
  string // Return of join
>(
  R.join('\n'),
  R.map(x => `${x[0]}: ${x[1]}`),
  R.toPairs
);

Personally I prefer to use pipe as the typings match up to the argument order within pipe, compared to compose which is backwards:

const rawHeaders = R.pipe<
  { [key: string]: string | number }, //Argument
  Array<[string, string]>, // return of toPairs
  string[], // return of map
  string // return of join
>(
  R.toPairs,
  R.map(x => `${x[0]}: ${x[1]}`),
  R.join('\n')
);

Either way, each function in the pipe/compose gets the right value and you don't need to decorate the function within the pipe exclusively(unless you start using something like R.flip). It's verbose, but it works.

(you can specify as many args as you need for the first function, the overloads will handle the rest btw)