Scratch

Germanium_Tinide · 2011-08-17 17:57:55

I have noticed, that the [ersetze (x) Element von [Liste] durch (y)] block ("replace item (x) of [List] with (y)") works only when there already are x Elements in the List... sad

Is there a Way to bypass this Problem (for example, if I want to set Item 2 to something regardless of whether or not it is there already), without using a long if-else Sequence or adding x Items and then deleting them after the Replacement)? Both of those Alternatives are tedious and do not always work... hmm

There are some other Problems with Lists too... I will talk about them later.

Harakou · 2011-08-17 18:19:24

Do you mean like a list has to be at least x items long for the "replace item x" block to work? Yes, I suppose that could be a problem.

You could use a loop to check if the list is x long and add items until it is, but you don't seem to want to do that. Other than that I see no way to do it though. hmm

Code:

Hat block of choice
other code
if (length of list y < x)
    repeat until (length of list y = x)
        add "null" to list y
replace item x of list y with z

Scratch

archived forums

#1 2011-08-17 17:57:55

A Problem with using Lists...

#2 2011-08-17 18:19:24

Re: A Problem with using Lists...

Code:

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