Re: Table.remove() issue with the last element

ok, very clear : thanks for the explanation.

Bye

Laurent

Le Mardi 21 juin 2016 0h17, Doug Currie <doug.currie@gmail.com> a écrit :

Laurent,

Don't replace table.remove(f, k), it's ok.

table.remove is meant to be used with tables with "valid indices." Your top-level table has "holes" and non-integer keys, whereas your deeper tables have "valid indices."

The other issue is that you are modifying tables while you are iterating over them. The combination of table.remove for a "list" iterated over with ipairs, and setting to nil with pairs seems to be ok, but not other combinations.

ipairs, remove with table.remove

pairs, remove with t[k] = nil

On Mon, Jun 20, 2016 at 6:07 PM, Laurent FAILLIE <l_faillie@yahoo.com> wrote:

Le Mardi 21 juin 2016 0h00, Doug Currie <doug.currie@gmail.com> a écrit :

> You should change
>
> table.remove(tbl, i)
>
> to
>
> tbl[i] = nil

But why can I do it for the top level table and not inner ones ?

xtbl = {
[1] = { "A" },
[2.5] = { "A", "B" },
[.5] = { "A" },
[5] = { "A", "B", "C" }
}

function dump(tbl)
    for i,f in pairs(tbl) do
        print(i, #f, f)

        for k,v in ipairs(f) do
            print(' ---', k, v)
        end
    end
end

function remove(tbl, func)
    for i, f in pairs( tbl ) do
        for k,v in ipairs(f) do
            if v == func then
                f[k] = nil
--                table.remove(f, k)
            end
        end
        if #f == 0 then
print('** remove', i)
            tbl[i] = nil
--            table.remove(tbl, i)
        end
    end
end

dump(xtbl)
remove(xtbl, 'A')

print("Resultat")
dump(xtbl)

1    1    table: 0x7bf4a8
---    1    A
5    3    table: 0x7c00b0
---    1    A
---    2    B
---    3    C
2.5    2    table: 0x7bf3c8
---    1    A
---    2    B
0.5    1    table: 0x7bf1d0
---    1    A
** remove    1
** remove    0.5
Resultat
5    3    table: 0x7c00b0
2.5    2    table: 0x7bf3c8

As you can see, I can't walk against inner ones :(