Infer function argument type in typescript

Question

I am trying to automatically infer function argument types as follows:

const FUNCTION_A = "a";
const FUNCTION_B = "b";

type FunctionKeyType = typeof FUNCTION_A | typeof FUNCTION_B;

function functionA(arg: string): string {
    return "A";
}

function functionB(arg: number): string {
    return "B";
}

type AnyFunctionType = (arg: any) => string;

const funcs: Record<FunctionKeyType, AnyFunctionType> = {
    [FUNCTION_A]: functionA,
    [FUNCTION_B]: functionB
}

console.log("A:", funcs[FUNCTION_A]("a")) // works
console.log("B:", funcs[FUNCTION_B](0))  // works
console.log("B:", funcs[FUNCTION_B]("fail")) // should fail typechecking but does not

There will be a large number of functions with unique arg types, but each will always have a single argument called arg and return a string
I need to have a mechanism for key => function, however it doesn't necessarily need to be a Record

Is there a way to write AnyFunctionType such that when I use a function via funcs["..."], it infers the type of the argument?

TS Playground

Will the calls to the functions be hardcoded like that? funcs["a"]("a")? Or will the key be a plain string? — T.J. Crowder
– T.J. Crowder, Commented Apr 28, 2022 at 16:57
It will use typed string literals for keys as well as for accessing the functions (Edited post to reflect key types) — krames
– krames, Commented Apr 28, 2022 at 16:59

Alex Wayne · Accepted Answer · 2022-04-28 17:01:16Z

When you declare the type as Record<string, AnyFunctionType> then all functions are assumed to be of type AnyFunctionType which takes any as an argument.

If you want to apply a constraint, but infer something more specific than that, then you need to use a generic function to make this object. (See issue#47920 for how this might be solved in the future)

function makeFuncs<T extends Record<string, AnyFunctionType>>(funcs: T) {
    return funcs
}

const funcs = makeFuncs({
    "a": functionA,
    "b": functionB
})

const badFuncs = makeFuncs({
    "a": functionA,
    "b": functionB
})

console.log("A:", funcs["a"]("a")) // works
console.log("B:", funcs["b"](0))  // works
console.log("B:", funcs["b"]("fail")) // error

And it enforces that the the function signature is correct:

const badFuncs = makeFuncs({
    "a": functionA,
    "b": () => 123 // type error
})

Playground

Bishwajit jha · Accepted Answer · 2022-04-28 17:06:59Z

1

You can make use of a Generic Function to implement this constraint.


type AnyFunctionType = (arg: any) => string;


function InferFunctionType<T extends {[k: string]:  AnyFunctionType }>(t: T){
    return t
}

const funcs =  InferFunctionType({
    "a": functionA,
    "b": functionB, 
})



console.log("A:", funcs["a"]("a")) // works
console.log("B:", funcs["b"](0))  // works
console.log("B:", funcs["b"]("fail")) // Here Typechecking gives error

Code Playground

answered Apr 28, 2022 at 17:06

Bishwajit jha

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Comments

T.J. Crowder · Accepted Answer · 2022-04-28 17:00:26Z

0

You've said the calls will be as shown, where the function name is a compile-time constant. That's good news! That means you can do it. :-) You can have a compile-time constant object to do the lookup on:

const funcs = {
    "a": functionA,
    "b": functionB
} as const; // <== Note

console.log("A:", funcs["a"]("a"));         // Works
console.log("B:", funcs["b"](0));           // Works
console.log("A wrong:", funcs["a"](0));     // Error as desired
console.log("B wrong:", funcs["b"]("x"));   // Error as desired

Playground link

You get full IDE support there as well, the intellisense tells you the argument type required.

answered Apr 28, 2022 at 17:00

T.J. Crowder

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2 Comments

Bishwajit jha Over a year ago

@t-j-crowder, but this won't stop us from assigning anything to funcs right? like const funcs = { a: () => Array<[]>, b: boolean } as const because nothing is constraining us from doing that, which is required in the problem statement I guess

T.J. Crowder Over a year ago

@Bishwajitjha - Yes, it won't. No, I don't think that's required by the problem statement. :-) But it's up to the OP (and others using our answers) to decide for themselves what's appropriate for them.

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