removing sprites from tables with a for

ricvieira · October 2013

hi all, just a quick question, I have a table with some sprites there, called birdList, if I iterate it from back to front, I can iterate it all and remove all like this:

count= #birdsList
for i=count, 1,-1 do
table.remove(birdsList, i)
end

but the same list, if i iterate it from start to end i get an error, that one of them, the 6th is nil (i have 10 sprites there)

count= #birdsList
for i=1, count do
table.remove(birdsList, i)
end

I've searched the forum for an answer to this behaviour in the for cycle, i couldnt find any, can someone explain to me whats happening

tnx in advance and sry about the noob question

ar2rsawseen · October 2013

Yes the problem is that once you remove the element from table, tables count becomes lesser than it is defined in your loop

The solution is to loop backwards

for i=#birdsList,1,-1 do
    table.remove(birdsList, i)
end

or control loop counter:

local i=1
while i <= #birdsList do
        table.remove(birdsList, i)
end

hope that helps

ricvieira · October 2013

yeah, it helps a lot, tnx

I thought I was controling the loop counter, with the

count= #birdsList
for i=1, count do

but from what you're saying is the variable "count" is being changed by removing an item from the list, even if its outside the for cycle, right?

again, sry about the noob questions, but im writting this down so I wont forget lol, lua and its global scope is very different from what Im used to programming

tnx again

ar2rsawseen · October 2013

No count remains the same, while the real size of table is getting smaller, so you finally reach the index of element that does not exist anymore, because you are using old size of table

OZApps · October 2013

@ricvieira,

 tbl = {} -- has 0 elements
 table.insert(tbl, "ONE") -- has 1 element
 table.insert(tbl, "TWO") -- has 2 elements
 table.insert(tbl, "THREE") -- has 3 elements
 
 --[[
Now the tbl has 3 elements and they are like an index, which is dynamic, so there is no guarantee that the element at 1 will be 1, unlike how it is with VB or Java.
 
so when you use table.remove(tbl, 2) , yes element 2 is removed and... HERE's the important bit,
 
tbl[1] is still "ONE"
and tbl[2] is now "TWO"
and there is no tbl[3] 
 
but you would ask what happened to tbl[3], I removed tbl[2] not [3]. That is the reason *why* a reverse loop is used, so that when the table dynamically reassigns the indexes, you do not encounter an error
--]]
 
do if you used
for i=#tbl, 1, -1 do
 table.remove(tbl, i)
end
 
-- it will work perfectly fine, but if you had it the other way... errors
 
-- NOTE: that if the table hasd key value pairs, they would not get reassigned automatically and #tbl would not include them in the result.

If you need to spend a little more time understanding Lua, try @Ar2rsawseens's site or my blogs at http://howto.oz-apps.com and http://learnlua.oz-apps.com

ricvieira · October 2013

tnx ozapps and yeah, ive been using your blog and ar2rsawseen site, great sites by the way, congrats

just sometimes, small details arent on the tutorials, thats why i asked here

tnx again

-- NOTE: that if the table hasd key value pairs, they would not get reassigned automatically and #tbl would not include them in the result.

yep, thats what was happening to me, its a table with key value pairs, it was a noob mistake lol

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