Even more? parse
[1/2] from: Rebolinth::nodep::dds::nl at: 29-Jul-2003 17:31
That was a wrong lefthanded hit...
Still not working.. I though I had it because of the spaces missing from the
initial string to search from but that was not the issue..
the result I would like -> "this is a long string" "long" is
[ "this" "is" "a" "" string" ]
Smash me, to let me understand it for now...
Norman.
[2/2] from: tomc:darkwing:uoregon at: 29-Jul-2003 17:02
parse has different modes depending on its arguments
when both arguments are string! the effect is to return a block!
with the first string split by the chars of the second string.
the typical use ie to turn a string into a block of strings
>> parse string none
== ["this" "is" "a" "long" "string"]
>> parse string "s"
==["thi" "i" "a" "long" "" "tring"]
broken up by whitespace and the char! s
the /all refinement makes whitespace significant
>> block: parse/all s "s"
== ["thi" " i" " a long " "tring"]
>> mask: unique "thisisalongstring"
== "thisalongr"
delimiters are not included in the output
>> parse string mask
== ["" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""]
perhaps the main way parse is used is with a string and a block of rules
in this mode parse returns true or false depending on whether the rules
were sufficent to get you to the end of the string.
>> parse s [to "long" "long" to end]
== true
>> cs: charset mask
== make bitset! #{
00000000000000000000000082D31C0000000000000000000000000000000000
}
>> parse string [some cs]
== true
within the rules block you use parens to do stuff in rebol instead of
in parse ... words heve differnt meanings within the .. dialect of parse
>> parse s [to "long" (print "hello") "long" to end]
hello
== true
result: copy ""
parse string [
copy token to "long" (append result token)
"long"
copy token to end (append result token)
]
== true
>> result
== "this is a string"
out of time... hope that helps
On Tue, 29 Jul 2003, Rebolinth wrote: