Joining Lists in Objective-C

Creation QR Code 2d barcode in Objective-C Joining Lists

You join lists together with the concatenation operator, &. Whenever you use the concatenation operator to combine two lists, you get a single list as a result, which is made of the items from the list to the left of the operator followed by the items from the list to the right. Script 6-1 shows a few examples of list operations.

Script 6-1. set new_list to {1, 2, 3} & {4 ,5 ,6 }--> {1, 2, 3, 4 ,5 ,6} set list_1 to 1 & 2 & 3 & 4 & 5 & 6--> {1, 2, 3, 4 ,5 ,6} set list_2 to "a" & "b" & "c" & "d" --> "abcd" Oops . . . what happened here list_1 ended up as a list, but list_2 ended up as a string. Remember that the concatenation operator works on strings as well as on lists, and if the left operand in a concatenation is a string, then the result will be a string as well. What you can do is turn the first item into a list, like this: set list_2 to {"a"} & "b" & "c" & "d" This gives you the result {"a", "b", "c", "d"} that s better.

Once a list contains one or more items, you can set the value of any item in a list to a different value, like this: set item n of the_list to new_item For example: set my_list to {"A", "B", "C"} set item 2 of my_list to "Z" my_list --> {"A", "Z", "C"}

Once you have either built a list or returned a list as a result, you need to be able to extract items from it. The example in Script 6-2 creates a list with six items, assigns it to the variable my_list, and then extracts items from it using various references. Script 6-2. set my_list to {1, 2, 3, 4 ,5 ,6} item 3 of my_list --> 3 first item of my_list -- Or "item 1 of my_list" --> 1 last item of my_list -- Or "item 1 of my_list" --> 6 middle item of my_list --> 3

You can also use some item to identify a random item in a list, as follows: set winner_name to some item of lottery_entry_list You can also get a range of list items from the same list using the thru keyword: items 2 thru 5 of {1, 2, 3, 4 ,5 ,6} --result: {2, 3, 4 ,5} items 2 thru 1 of {"a", "b", "c", "d", "e"} --result: {"d", "e"} If the script tries to access an item in a list that is out of range, the script will generate a Can t get reference error (error 1728), which is a common runtime error: item 4 of {"a", "b", "c} --error 1728: can't get item 4 of {"a", "b", "c} Getting a range of items from a list is also useful if you want to delete an item in a list, because AppleScript doesn t provide a way to delete list items directly. Instead, you have to construct a new list containing all the items except the one you don t want. For example, to delete the third item of a five-item list, you get a sublist containing all the items to the left of item 3 and another sublist containing all the items to the right of it and then join these two lists to produce the final four-item list: set the_list to {"a", "b", "c", "d", "e"} set new_list to (items 1 thru 2 of the_list) & (items 4 thru 5 of the_list) new_list --> {"a", "b", "d", "e"} Here s an example of how you d use this principle in a script: set the_list to {"a", "b", "c", "d", "e"} set remove_item to 3 set new_list to (items 1 thru (remove_item 1) of the_list) & (items (remove_item + 1) thru 1 of the_list) new_list --> {"a", "b", "d", "e"} This code isn t quite ready for general use, however. Although it works for any item in the middle of the list (2, 3, or 4), it will raise a Can t get reference error for the first or last item of the list as it tries to get an item that s out of range: set the_list to {"a", "b", "c", "d", "e"} set remove_item to 1 set new_list to (items 1 thru (remove_item 1) of the_list) & (items (remove_item + 1) thru 1 of the_list) --error: Can't get items 1 thru 0 of {"a", "b", "c", "d", "e"}. So you have to add a conditional statement that checks where the item is in the list and does things differently if it s at the start or end. Check out Figure 6-3 for how you can do that.