Empty object in ternary operator produces invalid type

[olehmisar]

Bug Report

🔎 Search Terms

empty object, {}, inferred type

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ

⏯ Playground Link

Playground link with relevant code

💻 Code

const x = true ? 'I am a string' : {} // type of `x` is: {}

🙁 Actual behavior

Type of x is inferred as {}

🙂 Expected behavior

Type of x should be inferred as "I am a string" | {}.

For example, if we try this code:

const x = true ? 'I am a string' : 5

...the type of x is inferred as "I am a string" | 5

[MartinJohns]

Strings are valid {}, so the type gets simplified. This is working as intended.

[jcalz]

See https://www.typescriptlang.org/docs/handbook/type-inference.html#best-common-type

[olehmisar]

Strings are valid {}, so the type gets simplified. This is working as intended.

why isn't this code simplified?

type X = string | {} // `X` is string | {}

[RyanCavanaugh]

Subtype reduction isn't required to happen, but is allowed to happen at any point. There's not a bug here.

[Jack-Works]

but do you think this type reduction is correct / meaningful to developers?

[poksme]

I encountered this issue when composing a dynamic object. For anyone seeing this and trying to find a workaround, couple things to note :

• Subtype reduction happens when assigned to a variable, so you might get away with inlining this code instead of storing it in a variable.

(true ? 'I am a string' : {}) // Type is 'I am a string' | {}
const myVar = true ? 'I am a string' : {} // Type is {}

• In my case, both return values of the ternary were enumerable, and cloning it with a spread prevented the subtype reduction, even when assigned to a variable.

const computeData = () => ({ foo: "bar"})

const subtypeReduced = true ? computeData() : {}; // Type is {}
const preservedType = {...(true ? computeData() : {}) } // Type is { foo?: string | undefined; }