Can I call a list inside of a list like this?

ID:140583
 
Nov 14 2009, 5:53 pm
Code:
M.PermVars["JutsuRef[jutsu]"] = "Not Known"


Problem description:
Basically, jutsu is a string in the JutsuRef list, this is making it so it shows that jutsu is not known in the reference page.

Can I call a string inside of a list inside of a list like this?
Nov 14 2009, 5:55 pm
Yeah, just don't put quotes around the list variable.
Nov 14 2009, 5:56 pm
In response to Kaiochao
Er, so is the format I used fine? Or will I need to add definitions so I don't use quotations?
Nov 14 2009, 5:58 pm
In response to Asellia
 
M.PermVars[JutsuRef[jutsu]] = "Not Known"
Nov 14 2009, 5:58 pm
In response to Kaiochao
That gives
Code\Figure Positions.dm:11:error: JutsuRef: undefined var


As an error?
Nov 14 2009, 6:54 pm
In response to Asellia
Let me guess, your trying to get a item out of the JutsuRef list?

Nov 15 2009, 6:08 am
In response to Asellia
You said that JutsuRef is a list. If it really is, you won't get an error saying it's not defined. If it's not defined, you don't really have a JutsuRef list.
proc/example()
    var/list
        L1[0]
        L2[0]

    L1["test1"] = 0
    L1["test2"] = 5

    L2["hi"] = "test1"
    L2["bye"] = "test2"

    world << L1[L2["hi"]] // outputs 0
    world << L1[L2["bye"]] // outputs 5

    L1[L2["hi"]] = 10

    world << L1["test1"] // outputs 10
    world << L1[L2["test1"]] // outputs 10

    L2["hi"] = "blah"
    L2["now this goes to test1"] = "test1"

    world << L1[L2["hi"]] // outputs nothing, as L1[L2["hi"]] now returns null, as there is no "blah" in L1
    world << L1[L2["now this goes to test1"]] // outputs 10

You have to remember, if you're using an associative list, you have an element mapping to another value. In this case (as is common) you have text strings mapping to something. So L[text string here] returns whatever the text string is mapped to.

So, if you have L["a"] = 1, then any time you have L["a"] it will evaluate to 1. world << L["a"] + 2 will output 3.

Now, if L2["someText"]="a", then L2["someText"] will evaluate to "a" so world<<L2["someText"] outputs a. But since L["a"] is 1, L[L2["someText"]] will end up being L["a"] (because the L2["someText"] part is "a"), and the L["a"] will then be 1. So L[L2["someText"]] is 1.