Weird infer by const modifier in spread arguments

[danni-boii]

Bug Report

The const modifier does not behave as expected. Typescript would warn me to infer the const generic parameters when I use the spread argument in a function when the literal type object consists of 2 different primitive literals.

And I am not sure whether this type of behaviour is intentional.

🔎 Search Terms

const modifier, typescript5, spread syntax, spread argument

🕗 Version & Regression Information

This behaviour begins starting at typescript 5.0 (where the const modifier is introduced)
I have been seeing this behaviour between typescript 5.0 to the nightly build of 5.1

⏯ Playground Link

Playground link with relevant code

💻 Code

type A = 1 | '2';

function concat<const U>(...items: (A | U)[]) {
  return [...items];
}

const originalArray: A[] = [1, '2'];
const perfectConcatenatedArray = concat(...originalArray, 3, 4 ,5);
const okayConcatenatedArray = concat(...originalArray, '3', '4', '5');
const concatnatedArray = concat(...originalArray, 3, 4, 5, '6', '7');

🙁 Actual behaviour

Typescript is inferring that the const U should be 3 instead of 3 | 4 | 5 | '6' | '7'
It prompts an error stating that Number 4 is not assignable to the parameter of type A | 3 for concatnatedArray.
While both perfectConcatenatedArray and okayConcatenatedArray are perfectly fine

🙂 Expected behaviour

I would expect concatnatedArray to have type 1 | '2' | 3 | 4 | 5 | '6' | '7'.
It is possible to do so if I wrote it in a class manner and chain the concat function. But I cannot do it within the same concat function.

Playground link to the class version of the issue

[RyanCavanaugh]

Inference will infer that it's OK to form a union of types from the same primitive literal family, but not different ones. See also
#32596 (comment)

[typescript-bot]

This issue has been marked 'Working as Intended' and has seen no recent activity. It has been automatically closed for house-keeping purposes.