In Red language, how to split a string using split, but also keep the delimiters as nessecary

Question

I want to split a string with the split, meanwhile the string contains the string used as delimiter which should not be treated as delimiter. I tried in this way as shown in the following code:

>> split {1 + 3 `to-string #"^(60)"`c} "`"
== ["1 + 3 " {to-string #"} {"} "c"] ;;I think it should be ["1 + 3 " {to-string #"^(60)"}"c"] 

to-string #"^(60)" used here is to avoid the appearance of "``" which would be referred as delimiter, but it fails. It seems that the ^(60) is evaluated to "```" and then is used as delimiter by split. So, what is the mechanism of the split in Red language? And how to split a string meanwhile keeping those delimiters that should not be treated as delimiter.

Answer 1

^(60) is a so-called codepoint form that gets loaded as a ` character.

>> "^(60)"
== "`"

If you want to avoid that, you should either escape it manually:

>> {1 + 3 `to-string #"^^(60)"` c}
== {1 + 3 `to-string #"^^(60)"` c}

Or use raw strings:

>> %{1 + 3 `to-string #"^(60)"` c}%
== {1 + 3 `to-string #"^^(60)"` c}

splitting it afterwards is trivial:

>> split %{1 + 3 `to-string #"^(60)"` c}% #"`"
== ["1 + 3 " {to-string #"^^(60)"} " c"]

---

In case you want to keep ` character there, then split won't cut it. You need something like Parse:

>> string: {1 + 3 `to-string #"`"` c}
== {1 + 3 `to-string #"`"` c}
>> parse string [collect [keep to " `" " `" keep to "` " "` " keep copy match to end]]
== ["1 + 3" {to-string #"`"} "c"]
>> parse string [collect some [keep copy _ to copy match [" `" | "` " | end] match]]
== ["1 + 3" {to-string #"`"} "c"]