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Discussion on: Typescript Odd Number Type

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ecyrbe profile image
ecyrbe • • Edited

Ok, challenge accepted. If you are not affraid of generic programming you can do Odd or Even type check without limitation :

type Number2String<T extends number> = `${T}`;
type IsStringEven<T extends string> = 
  T extends `${string}${0|2|4|6|8}` ? true : false;

type IsStringOdd<T extends string> = 
  T extends `${string}${1|3|5|7|9}` ? true : false;

type EvenNumber<T extends number> = 
  IsStringEven<Number2String<T>> extends true ? T : never;

type OddNumber<T extends number> = 
  IsStringOdd<Number2String<T>> extends true ? T : never;

function guarantyOddNumber<T extends number>(n: EvenNumber<T>) {
  return n+1;
}

function guarantEvenNumber<T extends number>(n: OddNumber<T>) {
  return n+1;
}

guarantyOddNumber(6876546); // succeed
guarantyEvenNumber(6876546); // fail
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check on playground

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tylim88 profile image
Acid Coder • • Edited

good solution, I further simplified your code

type IsEven<T extends number> = 
  `${T}` extends `${number}${0|2|4|6|8}` ? T : never;

type IsOdd<T extends number> = 
  `${T}` extends `${number}${1|3|5|7|9}` ? T : never;

function guarantyOddNumber<T extends number>(n: IsOdd<T>) {
  return n;
}

function guarantEvenNumber<T extends number>(n: IsEven<T>) {
  return n;
}

guarantyOddNumber(6876545); // succeed odd
guarantyOddNumber(6876546); // fail odd
guarantEvenNumber(6876546) // succeed even
guarantEvenNumber(6876545); // fail even
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generic

however there is fundamental different between your solution and my original solution

yours can only be used in function, the original solution is intended to assign to a variable, for example

const a:OddNumber = 1
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which is why it is a odd number type, not odd number check

I do however agree that the function one is more versatile