Cannot bracket-index into an object after checking it's existence on that object.

[h-joo]

Bug Report

Cannot bracket-index into an object after checking it's existence on that object.

The same failure was reported in #43651, but I couldn't find a separate discussion or a ticket for this.

🔎 Search Terms

• implicitly has an 'any' type
• bracket indexing

🕗 Version & Regression Information

Worked in 4.2.4. Starts to break since 4.3.0-beta

⏯ Playground Link

Playground link

💻 Code

interface A {}
function foo(a:A) {
 if ('abc' in a) {
 const z = a;
 // Element implicitly has an 'any' type because expression of type '"abc"' can't be used to index type 'A'.
 // Property 'abc' does not exist on type 'A'.
 return z['abc']; 
 }
}

🙁 Actual behavior

Compiler gives an error that you cannot index into type 'A', since 'abc' does not exist in type 'A'

🙂 Expected behavior

Compiler recognizes the existence check and does not give an error.

Observation

This seems to be caused by the improvement in the control flow analysis/type inference. Before the type of z in the below example was 'never' prior to TS4.3, but now the compiler recognizes z being type A

Question : I have one question though, is it by design that it's allowed to index into never types?

[MartinJohns]

See #34867.

[DanielRosenwasser]

Yeah but this worked before - Anders Hejlsberg (@ahejlsberg) sounds like we need to at least document this as a breaking change unless we believe it's a regression.

[ahejlsberg]

The current behavior seems right to me, but I'm not sure what caused the change.

[ahejlsberg]

Caused by #38610, which is specifically about not narrowing to never. As I said above, I think the current behavior is more desirable, but I suppose it is a breaking change. But really, anything relying on the old behavior seems suspicious given that you can index never with anything and not see an error. The issue that #38610 fixes is that never isn't a correct reflection of the in check. The check absolutely could succeed since pretty much anything is assignable to {}.