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Using JSX syntax I wrote a function that takes either a float or an integer as a parameter and returns a formatted string. I'd like any feedback for refactoring the code.

const moneyFormat = (float) => {
  let answer = [];
  let firstHalf;
  let secondHalf;

  // check if whole number
  if (float % 1 === 0) {
    float = Math.trunc(float);
    firstHalf = float.toString();
    secondHalf = '.00';
  } else {
    float = float.toFixed(2);
    const splitNumArr = float.split('.');
    firstHalf = splitNumArr[0];
    secondHalf = '.' + splitNumArr[1];
  }

  let everyThree = 0;
  // push chars into answer array
  for (let i = firstHalf.length - 1; i >= 0; i--) {
    answer.push(firstHalf[i]);
    everyThree += 1;
    // insert space for every three chars 
    if (everyThree % 3 === 0) {
      answer.push(' ');
    }
  }
  // reverse array and add secondHalf
  answer = answer.reverse().join('') + secondHalf;
  return answer;
}

console.log(moneyFormat(2310000.159897)); // '2 310 000.16'
console.log(moneyFormat(1600)); // '1 600.00'

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    \$\begingroup\$ I assume you (like me) are a beginner and are making your own code for the things you learn while doing so :) If you were doing this for a production server you would, of course, Google to find stackoverflow.com/questions/149055/…. Just chucking this link here for anyone coming from a search engine looking for a money formatter. \$\endgroup\$ Dec 23, 2017 at 6:41

1 Answer 1

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  • Bad variable names. dollars and cents would be more appropriate that firsthalf and secondhalf... I mean, c'mon, these variables don't represent "halfs" of anything. float isn't very descriptive either. Perhaps rawvalue would be better, and maybe result_buffer would be better than answer.
  • The entire code for whole numbers is unnecessary. When you run toFixed() on a whole number it becomes a float (string) anyway.
  • Rather than creating a new variable and arbitrarily swapping between let and const, why not just reassign float (or in my version, raw_value). You won't need it anymore after that anyway.
  • You might wanna consider casting it to a float in case a string is passed to the function, in which case your toFixed would fail.
  • It doesn't make a hue difference whether you declare your variables at the top or as needed when you're working on a small function, but be consistent. And if you do declare them all at the top, I would suggest doing it in a single statement for a small performance boost. In this case, I would personally just declare them as you go to keep the function concise.
  • When you're doing a backwards loop like that you don't need all three statements in the for loop. You could just do for (let i = firstHalf.length; i--;). That said, looping backwards and then reversing (or looping at all for that matter) makes for smelly code. You can just use a little regular expression to chunk the chars up: /.{1,3}/g
  • Don't reassign a variable just to return it, just return it. It's less work for the computer that way.

const moneyFormat = (raw_value) => {
  raw_value = parseFloat(raw_value).toFixed(2).split('.');
  let dollars = raw_value[0],
    cents = '.' + raw_value[1],
    format_buffer = dollars
      .split('').reverse().join('') // reverse it
      .match(/.{1,3}/g) // chunk it
      .map(c=>c.split('').reverse().join(''))
      .reverse(); // reverse it again
  return format_buffer.join(' ') + cents;
}

console.log(moneyFormat(2310000.159897)); // '2 310 000.16'
console.log(moneyFormat(1600)); // '1 600.00'
console.log(moneyFormat("1600"));
console.log(moneyFormat("234123.2134123"));

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    \$\begingroup\$ Uh, when I run your code snippet your numbers are reversed in each group of three, e.g. moneyFormat(1600) -> 1 006.00 \$\endgroup\$ Dec 23, 2017 at 6:29
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    \$\begingroup\$ @DarkMatterMatt - lolwut.. it's 2am why am i answering things on stackexchange... fixed it but it's a little more complicated than i would have liked.. hopefully someone else has a better idea \$\endgroup\$ Dec 23, 2017 at 6:36
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    \$\begingroup\$ It might be worth defining const reverse = str => str.split('').reverse().join('') \$\endgroup\$
    – Gerrit0
    Dec 26, 2017 at 22:28

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