Code: Select all
Procedure.l FindOneOf( *set.character, *string.character, start=0)
Define.Character *cTemp
#INDXBASE = 1
If *set And *string
If start
*string + (SizeOf(Character) * start)
EndIf
Repeat
*cTemp = *set
Repeat
If *string\c = *cTemp\c
ProcedureReturn (*cTemp - *set)+#INDXBASE
EndIf
*cTemp + SizeOf(Character)
Until *cTemp\c = 0
*string + SizeOf(Character)
Until *string\c = 0
EndIf
EndProcedure
;-
set.s = "abcdef"
string.s = "Hello"
Debug FindOneOf(@set, @string) ; should be 5
Debug FindOneOf(@set, @string, 3) ; should be 0
stringToSearch.s = "The quick brown fox jumped over the lazy dog."
Debug FindOneOf(@"aeiou", @stringToSearch) ; gives 2, which is the index at which the "e" is found at.
Debug FindOneOf(@"aeiou", @stringToSearch, 4) ; gives 5 (position of 'u' in set)
I couldn't find a fast way of doing this without having to do at least 4 PB calls and a few more inside the loops (mid, etc).
If it's not behaving properly, please let me know and I'll see what I can do.
In case you wonder, in simple words: It's a FindString() but, with a list of characters, the search is done on a per character basis and not on a per group basis.
To my understanding, this routine originally returned the index from the SET and not the STRING. It also began to count from 0 instead of 1, if you want 1-based results, do set #INDXBASE to 1 instead of 0. I hope PB's compiler is smart enough not to do a +0 at the end, let's hope so.
- In my posted code I use 1 though, because it's similar to what PB does.
If it was useful or you know how to optimize it, let me know
Part of my project requires mid-advanced routines such as this one, that's the reason I seem to be reinventing the wheel, as some people say.
Feedback appreciated.