noUncheckedIndexedAccess does not compile on incorrect possible undefined

[Magnuti]

Bug Report

🔎 Search Terms

• noUncheckedIndexedAccess

🕗 Version & Regression Information

• v4.9.4
• noUncheckedIndexedAccess: true
• This is the behavior in every version I tried, and I reviewed the FAQ for entries since v4.1.5

⏯ Playground Link

Playground link with relevant code

💻 Code

const numbers = [1, 2, 3];

if (numbers.length > 0) {
  const first = numbers[0];
  const doubleFirst = first * 2;
}

🙁 Actual behavior

The code does not compile since the compiler thinks first can be undefined.

🙂 Expected behavior

The compiler should notice that first must be of type number and not number | undefined since the if statement ensures that.

Of course the code compiles if we disable noUncheckedIndexedAccess, but I would like to keep it enabled.

[MartinJohns]

Essentially a duplicate of #38000 and/or #42909. noUncheckedIndexedAccess does not change anything about this.

You can access the first element and check that one for undefined.

[Magnuti]

@MartinJohns it seems redundant having check the first element for undefined. The array cannot contain undefined elements and the array contains elements by the if(numbers.length > +) statement, therefore the first element first cannot be undefined.

[WesleyYue]

@MartinJohns it seems redundant having check the first element for undefined. The array cannot contain undefined elements and the array contains elements by the if(numbers.length > +) statement, therefore the first element first cannot be undefined.

if (numbers.length > x) doesn't check if the var is undefined, only the length, you can have [undefined, undefined, undefined].

Regardless, I would agree the error should not throw here since numbers is clearly defined. I'm running into a bunch of variations of this where the var is clearly defined but typescript appends undefined liberally.

[Magnuti]

@WesleyYue are you sure that the array can contain undefined values? If I explicitly specify the type on the array the code below won't compile.

const numbers: number[] = [1, 2, 3, undefined];

Even if I do not specify the type on the array the result is the same.

const numbers = [1, 2, 3];
numbers.push(undefined)

[Magnuti]

if (numbers.length > x) doesn't check if the var is undefined, only the length, you can have [undefined, undefined, undefined].

@WesleyYue I mean yes if the array was not clearly defined, but as you say the array is clearly defined. The error should not be thrown so we agree on that

[MartinJohns]

Checking the length property does not narrow the type, that's what the issues I linked are about. I suggest checking the indexed value instead of the length property, or simply use a type assertion.

[fatcerberus]

For the record, it's entirely possible for this to happen at runtime:

let arr: number[] = [];
arr[1] = 42;
console.log(arr.length);  // 2 (i.e. arr.length > 0)
console.log(arr[0]);  // undefined!

It's debatable whether most code needs to be hardened against that (sparse arrays are evil), but seems worth considering, at least.