Issue found when defining hexadecimal color type: "The expression generates a union type that is too complex to represent (2590)"

Here is the type declaration I'm using:

type Hex = '0'|'1'|'2'|'3'|'4'|'5'|'6'|'7'|'8'|'9'|'A'|'B'|'C'|'D'|'E'|'F'|'a'|'b'|'c'|'d'|'e'|'f';
// ↓↓ Error: Expression produces a union type that is too complex to represent
type HexColor = `#${Hex}${Hex}${Hex}${Hex}${Hex}${Hex}`;

This code is written in Typescript version 5.2.2.

If you'd like to test this out, here is the Playground link:

Test it out on Typescript Playground!



I want to express my appreciation to @jcalz for providing his answer. Based on his response and this final result on the playground.

Answer №1

Your situation is incredibly intricate. When you prompt TypeScript to verify against a total of 22^6 possibilities, you're essentially asking it to consider 113,379,904 potential outcomes.

Answer №2

When it comes to TypeScript, there isn't a direct equivalent for the HexColor type. The limitation of union types is that they can only handle around 100,000 members, making it difficult to encompass all possible combinations beyond that threshold. While a three-character string of Hex characters fits within this limit (223 < 11,000), anything four characters or longer exceeds it (224 > 200,000).

There has been ongoing discussion about introducing support for regular-expression validated string types in TypeScript, as seen in the open issue at microsoft/TypeScript#41160. If this feature were to be implemented, representing a HexColor type would become feasible. In the absence of such functionality currently within the language, an alternate approach must be taken.

At present, the closest approximation involves forsaking a specific HexColor type and opting for a generic approach with HexColor<T>, serving as a constraint on a literal type input T. The aim is to ensure that T extends HexColor<T> if and only if T could have been a valid HexColor. Furthermore, the concept involves defining HexColor<T> to write a generic utility function hexColor(), accepting only valid hex color inputs and returning the input itself. Instead of declaring something like const y: HexColor = "012345", you'd use const y = hexColor("012345"). This general methodology follows:

type HexColor<T> = ⋯;
const hexColor = <T extends HexColor<T>>(h: T) => h;
const y = hexColor("012345"); // okay
const n = hexColor("012GFE"); // error, "021GFE" is not "0210FE"

An example showcases one way to define HexColor<T>, generating a six-character string literal where each character mirrors the corresponding Hex character from

T</code if available, or defaults to <code>"0"
otherwise. For instance, "021GFE" becomes "0210FE" due to an invalid "G", while "ABCDEFA" translates to "ABCDEF" by trimming surplus characters.

This phenomenon encapsulates a tail-recursive conditional type mechanism. The L type parameter acts as an accumulator maintaining a tuple's length equal to processed characters count so far. Conversely, the A type parameter serves as an accumulated output tracker. By subdividing T into its first component F and the remaining string R, recursion ensues based on these redefined parameters.

The subsequent demonstration subtly varies in presenting the hexColor helper function compared to previous descriptions:

const hexColor = <T extends string>(
  h: T extends HexColor<T> ? T : HexColor<T>) => h;

To circumvent circularity errors inherent in

T extends HexColor<T></code, the workaround entails restricting <code>T
to string initially before embracing
HexColor<T></code within <code>h
's type definition. Despite its unconventional appearance, the
T extends HexColor<T> ? T : HexColor<T>
conditional type adeptly infers string literals for evaluation against
HexColor<T></code, yielding either <code>T
or
HexColor<T></code based on success or failure.</p>
<p>A validation test then verifies this operational workflow:</p>
<pre><code>const y1 = hexColor("012345"); // okay
const y2 = hexColor("ABCDEF"); // okay
const y3 = hexColor("00ff00"); // okay

const n1 = hexColor("ABCDE"); // error!
// Argument of type "ABCDE" is not assignable to parameter of type "ABCDE0".
const n2 = hexColor("ABCDEFA"); // error!
// Argument of type "ABCDEFA" is not assignable to parameter of type "ABCDEF".
const n3 = hexColor("012GFE"); // error!
// Argument of type "012GFE" is not assignable to parameter of type "0120FE".
const n4 = hexColor("whatever"); // error!
// Argument of type "whatever" is not assignable to parameter of type "00a0e0".

In conclusion, successful calls yield expected outcomes, whereas faulty ones prompt informative error messages detailing the nearest valid input. Although clarity may waver for entirely incorrect entries such as "whatever," the system strives to deliver interpretative feedback beneficial to perplexed developers.

Explore Code on Playground

Answer №3

If you're curious, you have the option to verify a valid hexadecimal color with varying lengths using an index. Here's my approach to solving this challenge. I utilized a third-party library for numerical operations to streamline the process.

import type { Call, N } from 'hotscript'

type HexChar = 'A' | 'B' | 'C' | 'D' | 'E' | 'F' | 'a' | 'b' | 'c' | 'd' | 'e' | 'f';
type Digit = '0' | '1' | '2' | '3' | '4' | '5' | '6' | '7' | '8' | '9';
type Hex = HexChar | Digit;

type StripHash<T> = T extends `#${infer S}` ? S : never;
type ParseHex<
  T,
  Acc extends string = '#',
  Index extends number = 0,
> = T extends `${infer Char}${infer Rest}`
    ? Char extends Hex
        ? Rest extends ''
            ? Index extends 2 | 5 | 7
                ? `${Acc}${Char}`
                : never
            : ParseHex<Rest, `${Acc}${Char}`, Call<N.Add<Index, 1>>>
        : never
        : never;

type ValidateHex<T extends string> = ParseHex<StripHash<T>>;

This function validates any hexadecimal color code of three, six, or eight characters without the hash symbol. It functions by recursively analyzing the last character in the input. If no characters remain, it verifies the index length. If it meets the criteria, the character is added to reconstruct the original string type. Otherwise, it returns "never".

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