Typescript type inference with default parameter

Question

Simple example of what I am trying to do:

const simpleExample = <T>(v: T = 'cat') => {
  return v;
};

Error:

Type 'string' is not assignable to type 'T'.
  'T' could be instantiated with an arbitrary type which could be unrelated to 'string'.ts(2322)

The inference will work if I don't try to have a default parameter, so maybe the options is to wrap it in something that does the defaulting.. I don't understand why this doesn't work..

My actual code:

I want each of these Shape, Z, and Output to be infered from arguments and it works if I don't supply the default arg...

const createHooks = <
  Shape extends z.ZodRawShape,
  Z extends z.ZodObject<Shape, 'strict'>,
  Output extends z.ZodTypeAny
>(
  schema: Z,
  effect: (s: Z) => z.ZodEffects<Z, Output> = (s: Z) => s.transform((i) => i)
) => <Source, FProps extends { [k: string]: Record<string, any> }>({
  name,
  getInstanceKey,
  load,
  fieldProps,
}: {
  name: string;
  getInstanceKey: (s: Source) => string;
  load: (s: Source) => Z['_input'];
  fieldProps: FProps;
}) => {
 // ...
 return { }
}

Gets the error:

'Output' could be instantiated with an arbitrary type which could be unrelated to '{ [k in keyof addQuestionMarks<{ [k in keyof Shape]: Shape[k]["_output"]; }>]: addQuestionMarks<{ [k in keyof Shape]: Shape[k]["_output"]; }>[k]; }'

Which is frustrating, because I want any 'arbitrary type' to cause an error.

kaya3 · Accepted Answer · 2021-09-02 02:15:15Z

1

The problem is that your function could be called like simpleExample<number>(), in which case it promises to return a function of type (i: string) => number but does not. You can solve this by declaring two overload signatures: one with the type parameter, and one which will use the default argument with no type parameter.

function simpleExample<T>(f: (i: string) => T): (i: string) => T;
function simpleExample(): (i: string) => string;
function simpleExample(f = (i: string) => i) {
  return f;
};

Playground Link

answered Sep 2, 2021 at 2:15

kaya3

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3 Comments

exrhizo Over a year ago

Okay, this makes sense for the first part of my question, but I use the type parameter extensively in my function body and I don't want to explicitly set the return type, because it is complicated.. I know this is a very specific so maybe out of luck on SO I could imagine typescript just error simpleExample<number>() but what can you do..

exrhizo Over a year ago

I seem to be able to use type S = typeof schema; in my function scope, still investigating

exrhizo Over a year ago

I'll accept this answer if there are no more thoughts that arrive

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