World: r3wp

there  is front middle and bacl padding

pad list [","]

for a comma seperated list

Pekr: "I did not find an easier way, so I parse for E, then I distinguish 
the sign, the number -5 in above case, and then I compose the string 
:-)"
!> a: 123.456 reduce[to integer! a remainder a 1]
== [123 0.456000000000003]

Maybe the base for something better (dont know how easy that parsing 
is?)

dunno ... you simply has some computation, it returns the result 
... and result may be something like 1 / 100, and you suddenly end-up 
with 1E-2, now what to do with that?

split in integer and fraction, pad integer, do something with fraction, 
join them.

fraction could be rounded, then copy/part.

Petr: <I wrote generalised solution in the past.>
Is that published?

Eric Long also did some nice work on a format.r function....I'm trying 
to find out if it is still available to the public (I have a copy 
of it)

I have my own function currently for that, it is just I dont find 
1E-2 really usefully for anything, or just maybe form should be tweaked 
to form decimals in their full representation ...

http://www.nwlink.com/~ecotope1/index.r?

ncie, rebol indexes our index.r's :) and word-browser was great to 
find "remainder".

short enough? http://polly.rebol.it/test/test/snippets/epad1.r

Looks good, Volker.
One problem: more than 9 leading zeroes:
 epad1 0 10 0
** Math Error: Math or number overflow

(There is a much larger limit on trailing zeroes after the decimal 
sign)

tricky. hmm, money should have a bigger scope?

Nine *should* be enough for most people -- I didn't have a specific 
case in mind. I was just testing the limits.

testing is good :)

Yep, money is worth its -erm, punning here.. adds more digits. (uploaded)

14 leading zeroes is probably enough for anyone :-)
One bug?
 epad1 0.09 1 2
== "0.90"   ;; 10 times too large!

thanks. will look at that.

forgot to pad remainder to. so 0.09 * 100 -> 9 . should be 09 so 
that appending 0s works. but got another bug-description on linux. 
there it is fixed.

i need to make a function wich can take n optional arguments, n can 
be from 1 to many arguments any help

I think Ladislav has done some work on this.

Otherwise supply it as a block :)

i whas thinking about the block option it seems more acurrate

supply it as a block
 is the proper solution I think

btw - what you think of following? - normally we can access paths 
in several ways ... block/5 (fifth element), block/literal (literal 
element), block/(expression) (expressionn to evaluate). Nonexistant 
elements do return 'none. But try block/("some string") - it returns 
error. Now the question is, if this is correct/consistent with other 
ways of how we access block elements, or not?

Question: <Unix Time>..   I found several functions on the web that 
show you how to create a unix-timestamp from a rebol time (now). 
 However, I am looking for the ability to convert a unix-timestamp 
back into rebol-time.  I started writing my own function but I think 
that leap-years might mess me up.  Anyone have any thought on the 
matter?

!> now + to-time 1e6
== 23-Oct-2005/13:26:16+2:00

i guess you can refine that for conversion. a unix-timestamp is an 
integer based on some date?

1/1/1970 0:00:00

an then 32 bits for miliseconds or something like this?

The unix-timestamp could be any date/time.. so I think I have to 
work top down.. first find the year then month, day, hour, min, sec. 
    Kind of like building a binary number from a decimal.. top-down.

Precision (ms) is used/not used depending on the OS / config.

I have some MRTG/RRDTool files that I am parsing for certain data 
and they all use the unix-timestamp format (not including ms)

i thought it is just an integer. So it is more complicated build, 
with year, month etc?

Yeah.  the date -> unixTime function from someone else (maybe Gabrielle) 
is..
to-unix-time: func [date] [
  date/date - 1-1-1970 * 86400 + to-integer date/time
]

then you could reverse that. as my example shows, you can add times 
to dates. i think it works also in that large range.

Yeah.  My problem (I think) is accounting for leap years and the 
different lengths of months.

Little casio watches know all of this info (Year/month/day/time). 
 I suspect it is in a static table inside the watch.  I have been 
looking for something like this on the web.

I could easily build a table for month lengths.. it is the leap-year 
that I just don't want to think about.

That should be inbuild. i guess rebol stores internally just a big 
integer too.

you could try
  1-1-1970 + unix-seconds
and see what happens.

1-1-1970 + to-time unix-seconds

You might be on to something.. I did unix time (now) and it returned 
24-Dec-13007.  However, maybe I need to subtract not add.  I will 
play with this SIMPLE solution.  (I always try to look at it harder 
than it probably needs to be).  Thanks.

>> 1-Jan-1970/0:0:0 + to-time to-unix-time 15-Apr-2006/15:41:0
== 15-Apr-2006/15:41

Looks like the time just needs to be inlcuded...

i've this function: f: [variable] [variable: 100]
variable: 10
f variable
== 100
variable
== 10

the question is 1 why do i never noticed this :) and second how do 
i pass values by reference

You're not talking about this ?

 f: func ['word][ set word 10 ]

This might do it -- change the way you define f:
 use [variable] [f: [variable] [variable: 100]]
variable: 10
== 10
f variable
== 10
variable
== 10

note: "variable" is a global value still i get no results from this 
:(

>> f: func ['variable][ set variable 100 ]
>> variable: 10
== 10
>> f variable
== 100
>> variable
== 100