> I have received a bug report regarding AIList instances during save and load. See viewtopic.php?p=1279485#p1279485 > > I found out the reason the script crashes on load is due to a problem with saving in JGRPP. > > https://github.com/JGRennison/OpenTTD-p ... #L454-L455 > Between these two lines, adding the following fixes the save issue. > > > Code: > > > SlWriteByte(SQSL_INSTANCE); Thanks, this will be fixed in the next release. Statistics: Posted by JGR — 30 Jun 2026 23:46 * * *

I have received a bug report regarding AIList instances during save and load. See viewtopic.php?p=1279485#p1279485 I found out the reason the script crashes on load is due to a problem with saving in JGRPP. https://github.com/JGRennison/OpenTTD-p ... #L454-L455 Between these two lines, adding the following fixes the save issue. Code: SlWriteByte(SQSL_INSTANCE); Statistics: Posted by xarick — 30 Jun 2026 14:42 * * *

> LuDiAI is frequently crashing after a saved-game is loaded. I also note that version 26 is on GitHub, but only version 25 is in bananas. > Regards, > Habu Thanks for the bug report. I was only able to reproduce the crash on JGRPP, but not on OpenTTD 15. It seems there's an issue with the way saving / loading of ScriptList instances is handled. The crash happens when it is trying to get the Count of items of an AIList which is expected to exist at the time the script calls it, but it's not there. Saving and loading of ScriptList instances is a relatively recent feature of OpenTTD. I see it ported to JGRPP already, but something went wrong there. Statistics: Posted by xarick — 30 Jun 2026 11:21 * * *

> The postcondition you're quoting says that the list you copy from does not change as a result of the copy. Yes, the items are copied, not moved, so the source list remains identical. It does not say that the two list objects become "linked" or anything. Oh I totally misinterpreted the meaning, now I've done some new tests and yeah, AddList copies by value not by reference, indeed! IDK how I missed that initially when testing, but replacing array.extend with AIList::AddList works and it's better for what I need it, thank you, you're the best. > If you had object references as the data items on the list, the destination list would reference the same objects as the source list, but the list itself is still independent, so you could reorder, add, remove items from one list without affecting the other one. OK, I'm it looks still learning squirrel even though it feels like this is the only unknown to me, are you saying non-scalar items (e.g. array, instance, class) are by reference but the list or container itself is by value? meaning modifying the container's items is by reference while removing them from container is by value? I'll have to test this as well to be sure, there is not a single info online about how this works. Thank you clearing up! Statistics: Posted by GoldMine — 02 Mar 2026 09:58 * * *

AIList::AddList copies items from the list passed as parameter. If you have two source lists you add AddList them both to a single and get a new list with the items appended. The postcondition you're quoting says that the list you copy from does not change as a result of the copy. Yes, the items are copied, not moved, so the source list remains identical. It does not say that the two list objects become "linked" or anything. If you had object references as the data items on the list, the destination list would reference the same objects as the source list, but the list itself is still independent, so you could reorder, add, remove items from one list without affecting the other one. Statistics: Posted by jfs — 02 Mar 2026 09:18 * * *

> The postcondition you're quoting says that the list you copy from does not change as a result of the copy. Yes, the items are copied, not moved, so the source list remains identical. It does not say that the two list objects become "linked" or anything. Oh I totally misinterpreted the meaning, now I've done some new tests and yeah, AddList copies by value not by reference, indeed! IDK how I missed that initially when testing, but replacing array.extend with AIList::AddList works and it's better for what I need it, thank you, you're the best. > If you had object references as the data items on the list, the destination list would reference the same objects as the source list, but the list itself is still independent, so you could reorder, add, remove items from one list without affecting the other one. OK, I'm it looks still learning squirrel even though it feels like this is the only unknown to me, are you saying non-scalar items (e.g. array, instance, class) are by reference but the list or container itself is by value? meaning modifying the container's items is by reference while removing them from container is by value? I'll have to test this as well to be sure, there is not a single info online about how this works. Thank you clearing up! Statistics: Posted by GoldMine — 02 Mar 2026 09:58 * * *

AIList::AddList copies items from the list passed as parameter. If you have two source lists you add AddList them both to a single and get a new list with the items appended. The postcondition you're quoting says that the list you copy from does not change as a result of the copy. Yes, the items are copied, not moved, so the source list remains identical. It does not say that the two list objects become "linked" or anything. If you had object references as the data items on the list, the destination list would reference the same objects as the source list, but the list itself is still independent, so you could reorder, add, remove items from one list without affecting the other one. Statistics: Posted by jfs — 02 Mar 2026 09:18 * * *

> The postcondition you're quoting says that the list you copy from does not change as a result of the copy. Yes, the items are copied, not moved, so the source list remains identical. It does not say that the two list objects become "linked" or anything. Oh I totally misinterpreted the meaning, now I've done some new tests and yeah, AddList copies by value not by reference, indeed! IDK how I missed that initially when testing, but replacing array.extend with AIList::AddList works and it's better for what I need it, thank you, you're the best. > If you had object references as the data items on the list, the destination list would reference the same objects as the source list, but the list itself is still independent, so you could reorder, add, remove items from one list without affecting the other one. OK, I'm it looks still learning squirrel even though it feels like this is the only unknown to me, are you saying non-scalar items (e.g. array, instance, class) are by reference but the list or container itself is by value? meaning modifying the container's items is by reference while removing them from container is by value? I'll have to test this as well to be sure, there is not a single info online about how this works. Thank you clearing up! Statistics: Posted by GoldMine — 02 Mar 2026 09:58 * * *

AIList::AddList copies items from the list passed as parameter. If you have two source lists you add AddList them both to a single and get a new list with the items appended. The postcondition you're quoting says that the list you copy from does not change as a result of the copy. Yes, the items are copied, not moved, so the source list remains identical. It does not say that the two list objects become "linked" or anything. If you had object references as the data items on the list, the destination list would reference the same objects as the source list, but the list itself is still independent, so you could reorder, add, remove items from one list without affecting the other one. Statistics: Posted by jfs — 02 Mar 2026 09:18 * * *

> The postcondition you're quoting says that the list you copy from does not change as a result of the copy. Yes, the items are copied, not moved, so the source list remains identical. It does not say that the two list objects become "linked" or anything. Oh I totally misinterpreted the meaning, now I've done some new tests and yeah, AddList copies by value not by reference, indeed! IDK how I missed that initially when testing, but replacing array.extend with AIList::AddList works and it's better for what I need it, thank you, you're the best. > If you had object references as the data items on the list, the destination list would reference the same objects as the source list, but the list itself is still independent, so you could reorder, add, remove items from one list without affecting the other one. OK, I'm it looks still learning squirrel even though it feels like this is the only unknown to me, are you saying non-scalar items (e.g. array, instance, class) are by reference but the list or container itself is by value? meaning modifying the container's items is by reference while removing them from container is by value? I'll have to test this as well to be sure, there is not a single info online about how this works. Thank you clearing up! Statistics: Posted by GoldMine — 02 Mar 2026 09:58 * * *

I did a little test as follows and it appears copying an AIList is not by reference as documented, but by value: Code: function TestList() {local lst = AIList();lst.AddItem(0, 0);lst.AddItem(1, 1);lst.AddItem(2, 2);lst.AddItem(3, 3);lst.AddItem(4, 4);lst.AddItem(5, 5);lst.AddItem(6, 6);lst.AddItem(7, 7);local copy = AIList();copy.AddList(lst);lst.RemoveItem(0);Log.Info(lst.HasItem(0), lst.Count());foreach(item, v in lst) {Log.Info(item, v);}Log.Info(copy.HasItem(0), copy.Count());foreach(item, v in copy) {Log.Info(item, v);}} Attached image is the result: As you can see, the original list "lst" had item 0 removed, but the copy kept the item, however the docs HERE say this: > Postcondition > The list to be added ('list') stays unmodified. So what is going on? Isn't copy by reference? if it is then how is it possible that the item being removed stayed in copy, it should have been removed right? Is this just coincidence I shouldn't rely on? Please enlighten me. Statistics: Posted by GoldMine — 02 Mar 2026 05:20 * * *

AIList::AddList copies items from the list passed as parameter. If you have two source lists you add AddList them both to a single and get a new list with the items appended. The postcondition you're quoting says that the list you copy from does not change as a result of the copy. Yes, the items are copied, not moved, so the source list remains identical. It does not say that the two list objects become "linked" or anything. If you had object references as the data items on the list, the destination list would reference the same objects as the source list, but the list itself is still independent, so you could reorder, add, remove items from one list without affecting the other one. Statistics: Posted by jfs — 02 Mar 2026 09:18 * * *

I did a little test as follows and it appears copying an AIList is not by reference as documented, but by value: Code: function TestList() {local lst = AIList();lst.AddItem(0, 0);lst.AddItem(1, 1);lst.AddItem(2, 2);lst.AddItem(3, 3);lst.AddItem(4, 4);lst.AddItem(5, 5);lst.AddItem(6, 6);lst.AddItem(7, 7);local copy = AIList();copy.AddList(lst);lst.RemoveItem(0);Log.Info(lst.HasItem(0), lst.Count());foreach(item, v in lst) {Log.Info(item, v);}Log.Info(copy.HasItem(0), copy.Count());foreach(item, v in copy) {Log.Info(item, v);}} Attached image is the result: As you can see, the original list "lst" had item 0 removed, but the copy kept the item, however the docs HERE say this: > Postcondition > The list to be added ('list') stays unmodified. So what is going on? Isn't copy by reference? if it is then how is it possible that the item being removed stayed in copy, it should have been removed right? Is this just coincidence I shouldn't rely on? Please enlighten me. Statistics: Posted by GoldMine — 02 Mar 2026 05:20 * * *

I did a little test as follows and it appears copying an AIList is not by reference as documented, but by value: Code: function TestList() {local lst = AIList();lst.AddItem(0, 0);lst.AddItem(1, 1);lst.AddItem(2, 2);lst.AddItem(3, 3);lst.AddItem(4, 4);lst.AddItem(5, 5);lst.AddItem(6, 6);lst.AddItem(7, 7);local copy = AIList();copy.AddList(lst);lst.RemoveItem(0);Log.Info(lst.HasItem(0), lst.Count());foreach(item, v in lst) {Log.Info(item, v);}Log.Info(copy.HasItem(0), copy.Count());foreach(item, v in copy) {Log.Info(item, v);}} Attached image is the result: As you can see, the original list "lst" had item 0 removed, but the copy kept the item, however the docs HERE say this: > Postcondition > The list to be added ('list') stays unmodified. So what is going on? Isn't copy by reference? if it is then how is it possible that the item being removed stayed in copy, it should have been removed right? Is this just coincidence I shouldn't rely on? Please enlighten me. Statistics: Posted by GoldMine — 02 Mar 2026 05:20 * * *

Yes copy is by reference, but problem is that we're restricted to copy by reference. Only way to copy by value is to use array like shown, but array is not ideal when you're dealing with AIList or a tables resulting in code bloat and loss of ticks. So in my case I now have 15 lines of code bloat to have a copy by value. If there would be API to let copy by value it would take less ticks than having to do it manually. I though this is API restriction but it's true for tables as well, IDK why the language has such restriction. Statistics: Posted by GoldMine — 28 Feb 2026 18:09 * * *

Yes copy is by reference, but problem is that we're restricted to copy by reference. Only way to copy by value is to use array like shown, but array is not ideal when you're dealing with AIList or a tables resulting in code bloat and loss of ticks. So in my case I now have 15 lines of code bloat to have a copy by value. If there would be API to let copy by value it would take less ticks than having to do it manually. I though this is API restriction but it's true for tables as well, IDK why the language has such restriction. Statistics: Posted by GoldMine — 28 Feb 2026 18:09 * * *

For those who don't know what nasal demons are it's when you get frustrated with strange program flow so much you start to blame compiler, and out of frustration demons start to go out of your nose. (you can google for details if you want) I got frustrated because noticed squirrel doesn't let you copy AIList or tables. Actually you can copy them and your copy will be fine as long as original isn't modified, but as soon as you clear the source table or list your copy will no longer be a copy, data will be lost in both containers because for some unholy reason memory is somehow shared. Even if you manually copy AIList with AddItem it's still not going to be an independent copy, this is the most stupid language feature I've ever seen, why the f I can't make a copy? I found the only solution to make an independent copy of a container such that modifying the source leaves the copy alone is to create an array and extend it with data you want to copy. For instance: Code: local lst = [];local add = [];add.extend(lst); // I'm true independent copy The "add" is now true copy, you can clear or modify "lst" but add will keep the copied data. You can even remove elements from lst but copy keeps the data. Why can't this be done with AIList and tables? Why copying a list or table means having a copy that shares memory with source? This is really silly. It took me unreasonable time to figure out what's happening. Statistics: Posted by GoldMine — 28 Feb 2026 14:02 * * *

Yes copy is by reference, but problem is that we're restricted to copy by reference. Only way to copy by value is to use array like shown, but array is not ideal when you're dealing with AIList or a tables resulting in code bloat and loss of ticks. So in my case I now have 15 lines of code bloat to have a copy by value. If there would be API to let copy by value it would take less ticks than having to do it manually. I though this is API restriction but it's true for tables as well, IDK why the language has such restriction. Statistics: Posted by GoldMine — 28 Feb 2026 18:09 * * *

For those who don't know what nasal demons are it's when you get frustrated with strange program flow so much you start to blame compiler, and out of frustration demons start to go out of your nose. (you can google for details if you want) I got frustrated because noticed squirrel doesn't let you copy AIList or tables. Actually you can copy them and your copy will be fine as long as original isn't modified, but as soon as you clear the source table or list your copy will no longer be a copy, data will be lost in both containers because for some unholy reason memory is somehow shared. Even if you manually copy AIList with AddItem it's still not going to be an independent copy, this is the most stupid language feature I've ever seen, why the f I can't make a copy? I found the only solution to make an independent copy of a container such that modifying the source leaves the copy alone is to create an array and extend it with data you want to copy. For instance: Code: local lst = [];local add = [];add.extend(lst); // I'm true independent copy The "add" is now true copy, you can clear or modify "lst" but add will keep the copied data. You can even remove elements from lst but copy keeps the data. Why can't this be done with AIList and tables? Why copying a list or table means having a copy that shares memory with source? This is really silly. It took me unreasonable time to figure out what's happening. Statistics: Posted by GoldMine — 28 Feb 2026 14:02 * * *