Adding template literal key to spread object is ignored

[niedzielski]

🔎 Search Terms

"template literal" spread, spread "string literal"

🕗 Version & Regression Information

• This changed between versions ______ and _______
• This changed in commit or PR _______
• This is the behavior in every version I tried, and I reviewed the FAQ for entries about "spread"; present in nightly and goes back to at least v4.5.5.
• I was unable to test this on prior versions because _______

⏯ Playground Link

https://www.typescriptlang.org/play/?ts=5.8.0-dev.20250115#code/MYewdgzgLgBAbgQwE4EsECMA2BTAXDAAxQBMB9AEgG90UBzFMKAXwJgF4YByE0gRgCYAzJwDcAKFCRYEAA5JsCYuxiUAdOsoZg+MAFdMmJgBoYAbUSoMOALr4oSXdibiAFLPmKYEBFBQQAZijYECqmpPjQqGC0tjB6BkwAlGJAA

💻 Code

const variable: `id_${bigint}` = 'id_123';
const spread = {...{abc: null}, [variable]: true};
(spread satisfies {[_: string]: null})

🙁 Actual behavior

No errors. spread is incorrectly inferred as:

const spread: {
    abc: null;
}

🙂 Expected behavior

spread is inferred as:

const spread: {
    [_: `id_${bigint}`]: boolean;
    abc: null;
}

Causing the satisfies test to fail.

Additional information about the issue

Omitting the template literal errors correctly:

const variable = 'id_123';
const spread = {...{abc: null}, [variable]: true};
(spread satisfies {[_: string]: null})

Attempting to insert the template literal errors correctly:

const variable: `id_${bigint}` = 'id_123';
const spread: {[_: string]: null} = {...{abc: null}};
spread[variable] = true

[jcalz]

Effectively a duplicate of #38663. You're using a template literal key instead of a union, but the same thing happens: it is widened all the way to string as per #13948 ( #13948 (comment) ) and then it ends up disappearing in spread due ( #38663 (comment) )

[typescript-bot]

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