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Type guard is not working with noUncheckedIndexedAccess and destructuring #52623
Description
Bug Report
When noUncheckedIndexedAccess is on and you try to use a type guard like typeof list[index] !== undefined, it would not work if index was created via destructuring assignment or from the tuple.
🔎 Search Terms
noUncheckedIndexedAccess index destructuring undefined tuple
🕗 Version & Regression Information
- Compiler error
- Tested in 4.1.5, 4.5.5, and 4.9.5.
⏯ Playground Link
💻 Code
// index is a number
const index = 1
// indexD is a number
const [indexD] = [1]
const tuple: [number] = [1]
const indexT = tuple[0]
const list: Array<number> = [1, 2, 3]
// Block A
if (typeof list[index] === 'number') {
// `num` has a `number` type
const num = list[index]
}
// Block B
if (typeof list[indexD] === 'number') {
// `num` has a `number | undefined` type
const num = list[indexD]
}
// Block C
if (typeof list[indexT] === 'number') {
// `num` has a `number | undefined` type
const num = list[indexT]
}🙁 Actual behavior
Type guard does not work with the num in the Block B and C
🙂 Expected behavior
num in the Block B and C should have the same type as the num in the Block A
Activity
[-]Type guards are not working with `noUncheckedIndexedAccess` and destructuring[/-][+]Type guard is not working with `noUncheckedIndexedAccess` and destructuring[/+]MartinJohns commentedon Feb 5, 2023
Duplicate of #10530. Type narrowing does not occur for indexed access forms
e[k]wherekis not a literal.fatcerberus commentedon Feb 5, 2023
Workaround: Assign
list[index]to aconstfirst then type-guard that.PunGy commentedon Feb 6, 2023
I see... But still quite strange that even when you declare something as a constant literal, for example:
Very confusing behavior
typescript-bot commentedon Feb 16, 2023
This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.