Remind me again...

ID:150095
 
Dec 13 2001, 5:18 pm
In what case would a variable that displays as nothing still cause the statements (!v), (v == null), and (v == "") to be false? This variable was created with, essentially, var/v = copytext(msg,x,y) where msg is a regular old text string and both x and y turn out to be the same number. Or something like that. It's driving me nuts because I can't seem to screen it out afterwards by checking to see if it's null. And I could just check x and y before doing the copytext(), but I still want to know why all my checks with v are coming out false.

Z
Dec 13 2001, 6:28 pm
Zilal wrote:
In what case would a variable that displays as nothing still cause the statements (!v), (v == null), and (v == "") to be false?

!v is equivalent to: (v==null || v=="" || v==0)

The check (v==null && v=="") should never be true, since null and "" are two distinct values.

Does that answer the question?
Dec 14 2001, 3:28 am
Zilal wrote:
In what case would a variable that displays as nothing still cause the statements (!v), (v == null), and (v == "") to be false? This variable was created with, essentially, var/v = copytext(msg,x,y) where msg is a regular old text string and both x and y turn out to be the same number. Or something like that. It's driving me nuts because I can't seem to screen it out afterwards by checking to see if it's null. And I could just check x and y before doing the copytext(), but I still want to know why all my checks with v are coming out false.

Is it possible that v==" " or else v contains an HTML tag that doesn't display? The way I'd debug this would be something like:
if(v) world << "v=\"[v]\" (length=[length(v)])"

Lummox JR
Dec 14 2001, 9:47 pm
In response to Lummox JR
Lummox JR wrote:
Is it possible that v=="&nbsp;" or else v contains an HTML tag that doesn't display? The way I'd debug this would be something like:
if(v) world << "v=\"[v]\" (length=[length(v)])"


That had come to mind... I took your length debugging advice, and the length of v showed up as 0. The mystery survives...

Z
Dec 14 2001, 9:50 pm
In response to Tom
Tom wrote:
Zilal wrote:
In what case would a variable that displays as nothing still cause the statements (!v), (v == null), and (v == "") to be false?

!v is equivalent to: (v==null || v=="" || v==0)

The check (v==null && v=="") should never be true, since null and "" are two distinct values.

Does that answer the question?

No! Here's, like, what I did:

if (v==null) usr << "V is nothing!"

Upon running it, I found that v prints out as nothing, and has no length, so I changed the check to "if (v=="")", which had the same result, so I changed the check to "if (v==0)", which had the same result. It's like, somehow, v is a type of nothing other than null, "" and 0.

Z
Dec 14 2001, 11:35 pm
In response to Zilal
Zilal wrote:

No! Here's, like, what I did:

if (v==null) usr << "V is nothing!"

Upon running it, I found that v prints out as nothing, and has no length, so I changed the check to "if (v=="")", which had the same result, so I changed the check to "if (v==0)", which had the same result. It's like, somehow, v is a type of nothing other than null, "" and 0.

Can you get us a snippet? It's possible that I'm misunderstanding the situation in my late-night stupor, but it sure sounds like a bug to me.
Dec 15 2001, 10:01 pm
In response to Tom
Tom wrote:
Can you get us a snippet? It's possible that I'm misunderstanding the situation in my late-night stupor, but it sure sounds like a bug to me.

mob
verb
test(msg as text)
msg = Listall(msg)
for (var/v in msg) usr << "'[v]', length = [length(v)]"

proc
Listall(options)
var/list/outlist = new /list
var/colonplace
var/searchat = 1
do
colonplace = findtext(options,":",searchat,lentext(options)+1)
if (!colonplace || colonplace == 1) //only one choice to vote on... silly people!
outlist += options
return outlist

outlist += copytext(options,searchat,colonplace)
searchat = colonplace+1
if (searchat >= lentext(options)+1) return outlist

while (findtext(options,":",searchat,lentext(options)+1))
for (var/v in outlist)
if (!v) outlist -= v //supposed to remove null members

return outlist

Type in "test a:b:c::" to test.

Z
Dec 15 2001, 10:25 pm
In response to Zilal
The problem is that your proc is returning before it reaches the part where you filter out the null entries. Might I suggest an alternative?
proc
    Listall(options)
        var/list/outlist = list()
        var/colonplace
        var/searchat = 1     
        for()
            colonplace = findtext(options,":",searchat)
            var/v = copytext(options,searchat,colonplace)
            if(v) outlist += v
            if(!colonplace) return outlist
            searchat = colonplace+1
Dec 16 2001, 7:38 pm
In response to Tom
Tom wrote:
The problem is that your proc is returning before it reaches the part where you filter out the null entries.

I have no explanation for why I didn't think of that myself.

Might I suggest an alternative?
> proc
>   Listall(options)
>       var/list/outlist = list()
>       var/colonplace
>       var/searchat = 1     
>       for()
>           colonplace = findtext(options,":",searchat)
>           var/v = copytext(options,searchat,colonplace)
>           if(v) outlist += v
>           if(!colonplace) return outlist
>           searchat = colonplace+1  
>


I will in fact be able to take out the extra code I had in there to deal with situations in which text containing no colons was submitted... this whole thing is because one of the variables you have to submit in BYONDPoll had to not end in a colon, but people would forget and end it in a colon, so I'm trying to make it so you can do either... What does an argumentless for() statement do?

Z