Typescript recursive function composition

Question

I want to create a function chain, which would be an input of a pipe/flow/compose function.

Is this possible without the literal expansion of the types to selected depth, as is this usually handled? See lodash's flow.

I want to achieve typecheck of the data flow in the chain. - Argument of a function is result of the previous one - First argument is a template parameter - Last return is a template parameter

type Chain<In, Out, Tmp1 = any, Tmp2 = any> = [] | [(arg: In) => Out] | [(arg: In) => Tmp1, (i: Tmp1) => Tmp2, ...Chain<Tmp2, Out>];

The idea is in the draft.

This however produces tho following errors:

1. Type alias 'Chain' circularly references itself. (understand why, don't know how to resole)
2. A rest element type must be an array type. (probably spread is not available for generic tuples)
3. Type 'Chain' is not generic. (don't even understand why this error is even here)

Is this definition of Chain possible in Typescript? If so, please enclose a snippet.

(Tested on latest tsc 3.1.6)

Answer 1

Circular type aliases are not really supported except in certain cases. Instead of trying to represent the specific type you've written there in a TypeScript-friendly way, I think I'll back up and interpret your question as: how can we type a flow()-like function, which takes as its arguments a variable number of one-argument functions, where each one-argument-function return type is the argument type for the next one-argument-function, like a chain... and which returns a one-argument function representing the collapsed chain?

I've got something that I believe works, but it's quite complicated, using a lot of conditional types, tuple spreads, and mapped tuples. Here it is:

type Lookup<T, K extends keyof any, Else=never> = K extends keyof T ? T[K] : Else
type Tail<T extends any[]> = 
  ((...t: T) => void) extends ((x: any, ...u: infer U) => void) ? U : never;
type Func1 = (arg: any) => any;
type ArgType<F, Else=never> = F extends (arg: infer A) => any ? A : Else;
type AsChain<F extends [Func1, ...Func1[]], G extends Func1[]= Tail<F>> =
  { [K in keyof F]: (arg: ArgType<F[K]>) => ArgType<Lookup<G, K, any>, any> };
type LastIndexOf<T extends any[]> =
  ((...x: T) => void) extends ((y: any, ...z: infer U) => void)
  ? U['length'] : never

declare function flow<F extends [(arg: any) => any, ...Array<(arg: any) => any>]>(
  ...f: F & AsChain<F>
): (arg: ArgType<F[0]>) => ReturnType<F[LastIndexOf<F>]>;

Let's see if it works:

const stringToString = flow(
  (x: string) => x.length, 
  (y: number) => y + "!"
); // okay
const str = stringToString("hey"); // it's a string

const tooFewParams = flow(); // error

const badChain = flow(
  (x: number)=>"string", 
  (y: string)=>false, 
  (z: number)=>"oops"
); // error, boolean not assignable to number

Looks good to me.

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I'm not sure if it's worth it to go through in painstaking detail about how the type definitions work, but I might as well explain how to use them:

• Lookup<T, K, Else> tries to return T[K] if it can, otherwise it returns Else. So Lookup<{a: string}, "a", number> is string, and Lookup<{a: string}, "b", number> is number.

• Tail<T> takes a tuple type T and returns a tuple with the first element removed. So Tail<["a","b","c"]> is ["b","c"].

• Func1 is just the type of a one-argument function.

• ArgType<F, Else> returns the argument type of F if it's a one-argument function, and Else otherwise. So ArgType<(x: string)=>number, boolean> is string, and ArgType<123, boolean> is boolean.

• AsChain<F> takes a tuple of one-argument functions and tries to turn it into a chain, by replacing the return type of each function in F with the argument type of the next function (and using any for the last one). If AsChain<F> is compatible with F, everything's good. If AsChain<F> is incompatible with F, then F is not a good chain. So, AsChain<[(x: string)=>number, (y:number)=>boolean]> is [(x: string)=>number, (y: number)=>any], which is good. But AsChain<[(x: string)=>number, (y: string)=>boolean]> is [(x: string)=>string, (y: string)=>any], which is not good.

• Finally, LastIndexOf<T> takes a tuple and returns the last index, which we need to represent the return type of flow(). LastIndexOf<["a","b","c"]> is 2.

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Okay, hope that helps; good luck!