Typescript use parameter type inference with defaults

Question

I have this function as below that works well:

function pickAttributes<K extends string, T> (e: Element, attrs: K[]): Record<K, string> {
  return attrs.reduce((obj, key) => {
    return {
      ...(obj as any),
      [key]: e.getAttribute(key)
    }
  }, {})
}

I want to be able to have this function take an optional map function to convert the attributes to a different type.

function pickAttributes<K extends string, T> (e: Element, attrs: K[], mapFn?: (attr: null | string) => T): Record<K, T> {
  if (!mapFn) mapFn = x => String(x)
  return attrs.reduce((obj, key) => {
    return {
      ...(obj as any),
      [key]: mapFn(e.getAttribute(key))
    }
  }, {})
}

However i get this error

TS2322: 
Type '(x: string | null) => string' is not assignable to type '((attr: string | null) => T) | undefined'. 
  Type '(x: string | null) => string' is not assignable to type '(attr: string | null) => T'.
     Type 'string' is not assignable to type 'T'.

I get a similar error if i try to use a default parameter.

The changed function seems to work as expected without trying to add a default

function pickAttributes<K extends string, T> (e: Element, attrs: K[], mapFn?: (attr: null | string) => T): Record<K, T> {
  return attrs.reduce((obj, key) => {
    return {
      ...(obj as any),
      [key]: e.getAttribute(key)
    }
  }, {})
}

JB Nizet · Accepted Answer · 2018-07-22 11:32:22Z

1

First: I'm not sure what could extend string, so I would drop the generic parameter K.

But anyway: you want your return type to vary, based on the presence of the mapFn argument. I would solve that by defining two overloads of the function. And in the implementation, by using a union type string | T:

function pickAttributes<K extends string>(e: Element, attrs: K[]): Record<K, string>;
function pickAttributes<K extends string, T> (e: Element, attrs: K[], mapFn: (attr: null | string) => T): Record<K, T>;
function pickAttributes<K extends string, T> (e: Element, attrs: K[], mapFn?: (attr: null | string) => T): Record<K, string | T> {
  const fn = mapFn || (x => String(x));
  return attrs.reduce((obj, key) => {
    return {
      ...(obj as any),
      [key]: fn(e.getAttribute(key))
    }
  }, {});
}

answered Jul 22, 2018 at 11:32

JB Nizet

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1 Comment

megamit Over a year ago

Thanks the overload + union type seemed to do the trick. The reason why i use K extends string is because im using K as an enum. This way typescript knows that the return object has the strings in attrs as properties via Record

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