Type narrowing not working in else statements

[mon-jai]

Bug Report

🔎 Search Terms

• Narrowing
• Else statements

🕗 Version & Regression Information

Playground link with relevant code

💻 Code

type Foo = (
    { name: "a" | "b"; isAorB: true }
    | { name: "c"; isC: true }
)

declare let foo: Foo

if (foo.name == "a" || foo.name == "b") {
    console.log(foo.isAorB)
} else {
    const name = foo.name // TypeScript recognize that foo.name == "c" and therefore, name == "c"
    console.log(foo.isC) // Error reported, yet we know that foo.isC == true
}

🙁 Actual behavior

TypeScript failed to recognize foo.isC.

🙂 Expected behavior

TypeScript recognize that foo.isC == true.

[MartinJohns]

Duplicate of #43026.

[fatcerberus]

I’m not sure #43026 (comment) applies here since that involves a && in the initial if branch which can be decomposed into two separate (and insufficient) narrowing operations. The narrowings here should be atomic.

[MartinJohns]

Fairly certain it does apply. When foo.name == "a" is falsy, there's nothing the type of foo can be narrowed to. Same applies for the other expression. And when you remove the union it works fine:

type Foo = (
    | { name: "a"; isAorB: true }
    | { name: "b"; isAorB: true }
    | { name: "c"; isC: true }
)

[typescript-bot]

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