Passing a list item that is a list itself

Scripting Forums AppleScript | Mac OS X
#1

I’m sure this is pretty simple, but in the state of mind I’m in at the moment (I lost a LOT of data last night) the answer is evading me.

The following script works fine

set theList to {{"~/Desktop/", "Folder1"}, ¬
	{"~/Desktop/", "Folder2"}, ¬
	{"~/Desktop/", "Folder3"}}

repeat with i from 1 to count of items of theList
	set x to item i of theList
	MakeNew(x)
end repeat

on MakeNew({PathName, CreateName})
	do shell script "mkdir " & PathName & CreateName
end MakeNew

However this does not

set theList to {{"~/Desktop/", "Folder1"}, ¬
	{"~/Desktop/", "Folder2"}, ¬
	{"~/Desktop/", "Folder3"}}

repeat with i from 1 to count of items of theList
	MakeNew((item i of theList))
end repeat

on MakeNew({PathName, CreateName})
	do shell script "mkdir " & PathName & CreateName
end MakeNew

For the life of me I can’t figure out why this wouldn’t work. What’s the difference between passing the item of a list versus setting a variable to that item then passing the variable.

Thanks!

#2

In the second version, item i of theList is a list, but MakeNew specifies the contents of the list instead of just a list:

set theList to {{"~/Desktop/", "Folder1"}, ¬
	{"~/Desktop/", "Folder2"}, ¬
	{"~/Desktop/", "Folder3"}}

repeat with i from 1 to count of items of theList
	MakeNew(item i of theList)
end repeat

on MakeNew(myList)
	set {PathName, CreateName} to myList
	display dialog "OK"
end MakeNew

In the second version, this also works (strange, I know):

set theList to {{"~/Desktop/", "Folder1"}, ¬
	{"~/Desktop/", "Folder2"}, ¬
	{"~/Desktop/", "Folder3"}}

repeat with i from 1 to count of items of theList
	MakeNew(items of (item i of theList) as list)
end repeat

on MakeNew({PathName, CreateName})
	display dialog "OK"
end MakeNew
#3

Thanks Adam.

Well that both clarified and confused me even more. When you mention MakeNew looking for the contents of the list this line actually makes perfect sense to me:

MakeNew(items of (item i of theList) as list)

What confuses me though is since its looking for the contents I don’t quite grasp how this works:

set x to item i of theList
   MakeNew(x)

since isn’t it doing the same thing… passing the list rather then the lists contents?

#4

I’ll take a stab at this.

I believe that first iteration works because (item i) is 2 items, not a list, and the handler is expecting a 2 item list, so it takes the two items, and runs. The second iteration is sending a list (a single item), but the handler wants 2 items, so it does not fire. I think that is why Adam’s versions works; it is sending the items, not a list.

Of course, I have my preferences on how I would code this:

set theList to {{"~/Desktop/", "Folder1"}, ¬
	{"~/Desktop/", "Folder2"}, ¬
	{"~/Desktop/", "Folder3"}}

repeat with i in theList
	MakeNew(i)--Just send the list
end repeat

on MakeNew(pc)--since a list was sent, we know this is a list, without telling the handler explicitly
	do shell script "mkdir " & (pc's item 1) & (pc's item 2)--Separate the items here
end MakeNew
#5

Yeah I had that too, but was just trying to experiment with different ways to skin a cat. Thanks for the info though that makes a decent amount of sense =)

#6

Hi,

The reason it doesn’t work is that you’re passing a reference to a list item and not a list of two items. The same thing happens when you use the more common form of the repeat loop where this occurs.


set l to {{1, 2}, {3, 4}, {5, 6}}
repeat with this_item in l
	DoSomething(this_item)
end repeat
--
on DoSomething({x, y})
	display dialog (x + y)
end DoSomething

Note that here, you are using a variable in the call to the subroutine and it still doesn’t work because it’s a refernce to the list item.

gl,