The Typescript error condition will always result in a true verdict as there is no overlap between the 'string' and '0' types

After successfully converting a JavaScript number to words converter to TypeScript, I encountered an error in the block below

The issue arises as this condition always evaluates to 'true' due to the mismatched types of 'string' and '0'.

export const inWords = (num: any) => {
  if ((num = num.toString()).length > 9) return 'overflow';
  const n = ('000000000' + num).substr(-9).match(/^(\d{2})(\d{2})(\d{2})(\d{1})(\d{2})$/);
  if (!n) return;
  let str = '';
  str += n[1] !== 0 ? (a[Number(n[1])] || b[n[1][0]] + ' ' + a[n[1][1]]) + 'crore ' : '';
  str += n[2] !== 0 ? (a[Number(n[2])] || b[n[2][0]] + ' ' + a[n[2][1]]) + 'lakh ' : '';
  str += n[3] !== 0 ? (a[Number(n[3])] || b[n[3][0]] + ' ' + a[n[3][1]]) + 'thousand ' : '';
  str += n[4] !== 0 ? (a[Number(n[4])] || b[n[4][0]] + ' ' + a[n[4][1]]) + 'hundred ' : '';
  str +=
    n[5] !== 0
      ? (str !== '' ? 'and ' : '') + (a[Number(n[5])] || b[n[5][0]] + ' ' + a[n[5][1]]) + 'only '
      : '';
  return str;
};

I am unsure about resolving this particular error

javascripttypescript

Answer №1

The variable n is strictly for storing strings, regardless of whether num is a number, as seen in the concatenation on line 3

const n = ('000000000' + num)

Your ternary expressions involve a strict type check (!==) that examines both value and type.

Given that your value is consistently a string, it will never be equal to a number

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