CODEWITHSUNDEEP
lua
Mark Green
Mark Green

Reputation: 1330

Lua reports out of bounds differently based on size of a hole in a sequence

The following code:

myList = {1, 5, 9, 10}
myList[8] = 8
table.remove(myList,8)

Should produce a table with a border at 4, nil in keys 5 and 6, and 8 at key 8, then delete the 8. This program works.

However, the following:

myList = {1, 5, 9, 10}
myList[9] = 8
table.remove(myList,9)

does not work, saying that the call to table.remove is out of bounds!

This is contradictory: myList[9] is a non-null entry, and if table.remove requires that there are no borders before the given entry, then the border at 4 should have caused the first example to fail too. The only difference is that the gap in the keys is one step wider!

What's happening here and is it something I can correct?

Upvotes: 1

Views: 83

Answers (1)

Nicol Bolas
Nicol Bolas

Reputation: 473272

The part of a table that is an array (integer indices starting from 1 and ending at the first nil value) is the only part for which table.remove is valid. The array part cannot contain "holes"; the first "hole" represents the end of the array.

If you give table.remove indices that are not in the array part of the table, then you get undefined behavior. Maybe sometimes it will "work", and sometimes it won't. The workaround is to not specify indices outside of the array part of the table.

Upvotes: 3

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