Self referencing function in TypeScript?

Question

I'm wondering how to define the type for the defaultEncoder parameter. It has the same parameters as the function itself.

const paramsSerializer = (params: Params) => {
  const encoder = (
    str: string,
    defaultEncoder: any,
    charset: string,
    type: string
  ) => {
    const encodedStr = defaultEncoder(str, defaultEncoder, charset, type);
    return transformEncodedStr(encodedStr)
  };

  return transformParams(params, { encoder });
}

If I define the defaultEncoder every time it would be an infinite sequence of types.

defaultEncoder: (
  str: string,
  encoder: (
    str: string,
    encoder: // And so on...
    charset: string,
    type: "key" | "value"
  ) => any,
  charset: string,
  type: "key" | "value"
) => any,

It's easier if you extract the type: typescriptlang.org/play?#code/… — jonrsharpe
– jonrsharpe, Commented Apr 8, 2021 at 16:22
@jonrsharpe Thanks a lot. Didn't know that I can use the same type in the own type definition. Maybe you can post this as an answer so I can accept it. — Nepo Znat
– Nepo Znat, Commented Apr 8, 2021 at 16:30

0Valt · Accepted Answer · 2021-04-08 22:47:32Z

1

As jonrsharpe said in his comment you should extract the type to a type alias to recursively use it:

type Encoder = (
  str: string,
  defaultEncoder: Encoder,
  charset: string,
  type: string
) => string;

Then you can use it in the encoder function:

const paramsSerializer = (params: Params) => {
  const encoder = (
    str: string,
    defaultEncoder: Encoder, // here
    charset: string,
    type: string
  ) => {
    const encodedStr = defaultEncoder(str, defaultEncoder, charset, type);
    return transformEncodedStr(encodedStr)
  };

  return transformParams(params, { encoder });
}

To make it simpler you can use the Encoder type in the function definition itself:

const paramsSerializer = (params: Params) => {
  const encoder: Encoder = (str, defaultEncoder, charset, type) => {
    const encodedStr = defaultEncoder(str, defaultEncoder, charset, type);
    return transformEncodedStr(encodedStr)
  };

  return transformParams(params, { encoder });
}

edited Apr 8, 2021 at 22:47

0Valt

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answered Apr 8, 2021 at 16:31

blaumeise20

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3 Comments

jonrsharpe Over a year ago

If you do const encoder: Encoder = ... you don't need to reiterate all of the types of the parameters.

blaumeise20 Over a year ago

Yeah, that's also an option.

jonrsharpe Over a year ago

It's not just "an option"; it actually lets the compiler check the function is an Encoder, especially helpful as it includes the return type that's otherwise missed out.

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