Creating functions in a hierarchy where they get the same arguments

Question

When I write code, I tend to follow functional paradigms often. The main thing I try to do is use values returned from functions.

What I notice though, is I end up having a sort of function argument funnel where all the functions in the chain have the same arguments.


import fs from 'fs'
import path from 'path'

function dbFilePath(dbFilename: string) {
  const filePath = path.join(__dirname, "../", dbFilename + ".db")
  return filePath
}

export function fileData(dbFilename: string) {
  const file = dbFilePath(dbFilename)

  const fileContent = fs.readFileSync(
    file,
    "utf-8"
  );

  return fileContent
}

In this example code (which works fine), both functions have the same argument. This is an example of what I normally do.

What is best practice for structuring this piece of code?

Is the piece of code you posted something you'd actually used or heavily simplified? — Mast
– Mast ♦, Commented Apr 19, 2023 at 10:10
@Mast - this came directly from my source code. As it says, it works. — Vinn
– Vinn, Commented Apr 19, 2023 at 10:22

EzioMercer · Accepted Answer · 2023-04-20 10:24:20Z

You can create a function type which will accept only 1 parameter like this

type AccpetString<T> = (s: string) => T;

Then you need to save your functions in variables with type AccpetString<T> like this:

const dbFilePath:AccpetString<string> = dbFilename => {
  const filePath = path.join(__dirname, '../', dbFilename + '.db');

  return filePath;
}

or like this:

export const fileData:AccpetString<string> = function(dbFilename) {
  const file = dbFilePath(dbFilename);

  const fileContent = fs.readFileSync(
    file,
    'utf-8'
  );

  return fileContent
}

Now you don't need to specify the type of dbFilename in your functions

If your functions returns the string always, then we can declare a new type like this

type AccpetStringReturnString = AccpetString<string>

and rewrite your functions like this:

const dbFilePath:AccpetStringReturnString = dbFilename => {
  const filePath = path.join(__dirname, '../', dbFilename + '.db');

  return filePath;
}

or like this:

export const fileData:AccpetStringReturnString = function(dbFilename) {
  const file = dbFilePath(dbFilename);

  const fileContent = fs.readFileSync(
    file,
    'utf-8'
  );

  return fileContent
}

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