Inferred Type Predicates doesn't narrow type if using other conditional expressions

[zyoshoka]

🔎 Search Terms

"Inferred Type Predicates"

🕗 Version & Regression Information

5.6.0-dev.20240801

⏯ Playground Link

https://www.typescriptlang.org/play/?ts=5.6.0-dev.20240801#code/C4TwDgpgBA8lC8UDeUDWEQC4oGdgCcBLAOwHMoBfAbgCgBjAe2LygYCMArHbACjgB8oxAK4AbUQEoA2gF0EUKSnRYoAcgCGqygBpkaDNlVstFXSPEzajZsCgAzQqOAR8EACYBGeey4A6B04uPDgIAHy4UACEiOaSVFAA9AmwsvRMLAHOrm4ATN6cOP6OWcFhEdFCYqJQAGQ1uL7KCIgaqhLxSVB8UIKx0jJpNvbFLu4AzPl+mUEh8OEhFX1Fgfilcw1N8C2a7YnJMKlAA

💻 Code

type O = { key: string };
const objs: (O | null)[] = [{ key: 'a' }, { key: 'b' }, null];
const filtered1 = objs.filter(s => s != null); // O[]
const filtered2 = objs.filter(s => s != null && s.key == 'a'); // (O | null)[]
const filtered3 = objs.filter(s => s != null).filter(s => s.key == 'a'); // O[]

🙁 Actual behavior

The type inferred for filtered2 is (O | null)[] instead of O[].

🙂 Expected behavior

I would expect filtered2 to have the inferred type O[], similar to filtered1 and filtered3.

Additional information about the issue

No response

[MichalMarsalek]

This is working as intended. There are no one-sided typeguards.
The second case cannot be inferred as s is O, because if s.key != 'a', s can still be O.

[jcalz]

See #57465, where it mentions #15048

[zyoshoka]

Got it, thank you. I'm closing this issue.