After enabling noUncheckedIndexedAccess, using ++ on a default undefined does not trigger an error message. about typescript HOT 3 CLOSED

The linked issue is about what the language service (i.e. tooltips) reports about the left operand's type, not whether or not there's an error when you use ++ on that. It's not a duplicate.

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Sounds like a duplicate of #53303. edit: it's not.

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This problem also affects this form: hhh["asd"] = hhh["asd"]++;

I'm confused as to why the linked issue is closed as "Working as intended". The expression hhh["asd"]++ decomposes to:

let temp = hhh["asd"];
temp++;
hhh["asd"] = temp;

Written out that way, TS correctly identifies that we are reading a possibly-undefined value and then applying the ++ operator to it, which is an error. How is it correct that the same check should be elided with compound assignment when it represents identical logic?

This is the explanation offered:

Since it's both a read and a write, why is one of the answers correct and the other not? They have equal standing.

What has equal standing? The order of operations is read, increment, write. It's not one or the other, it's those three, in that order.

Edit: I guess that explanation is with respect to the type of the compound assignment expression—whether the type is inferred from the read side (number | undefined) or the write side (number). The linked issue is regarding the type of the expression. This issue is more about the failure to flag incrementing undefined as an error.

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