[Done] PB 6.40 Alpha 1 - Issue with referenced strings which are modified

[Matheos]

Small loop should be very fast, can you post your code ?

Yes, one example that concerns me especially, is where we have a console session open to provide an editor. The data is in a very proprietary form, which often contains control codes. We have to convert the control codes to a displayable symbol (not in the actual data but only when it's displayed).

Therefore, as the user moves around the data with PgUp/PgDn/Home/End and so on, we have to convert the control characters into a displayable character, normally 149 which is a dot, so the user can identify the characters that must not be changed or overwritten :

Code: Select all

ReplaceString(Params.s, Chr(60251), Chr(149), #PB_String_InPlace)

[Fred]

How big is param ? To me, using a PB function for this should be very close in term of speed. Something like (use it with caution, I didn't test it much):

Code: Select all

Procedure ReplaceChar(*String.Character, CharToSearch.c, CharToReplace.c)
  If *String = 0
    ProcedureReturn
  EndIf
  
  While *String\c
    If *String\c = CharToSearch
      *String\c = CharToReplace
    EndIf
    
    *String+1
  Wend
EndProcedure

Test$ = "Hello World !"

ReplaceChar(@Test$, 'o', 'O')

Debug Test$

With this proc, you can also replace several char in one pass.

[User_Russian]

Code: Select all

    *String+1

PB uses Unicode strings, and each character requires 2 bytes. I think it's better to do it this way.

Code: Select all

*String+SizeOf(Character)

[ChrisR]

If used with ReplaceChar(@Test$, 'o', ''), the length needs to be recalculated with PeekS
By changing the ReplaceChar procedure like this, you don't have to call PeekS
but Timo won't agree with this hack

Code: Select all

Procedure ReplaceChar(*String.Character, CharToSearch.c, CharToReplace.c)
  If *String = 0 : ProcedureReturn : EndIf
  
  Protected *Len.Integer = *String -SizeOf(Integer)
  *Len\i = 0
  While *String\c
    If *String\c = CharToSearch
      *String\c = CharToReplace
      If CharToReplace = 0
        Break   ; Comment it for testing the lenght 
      EndIf
    EndIf
    *Len\i + 1
    *String + SizeOf(Character)
  Wend
EndProcedure

Test$ = "Hello World !"
ReplaceChar(@Test$, 'o', 'O')
Debug Test$

ReplaceChar(@Test$, ' ', '')
Debug Test$
Debug Len(Test$)
;Test$ = PeekS(@Test$)
;Debug Len(Test$)

[STARGÅTE]

Interestingly, a self-written loop is faster than ReplaceString(, #PB_String_InPlace).
However, this self-written function only works for single character replacement.

Code: Select all

CompilerIf #PB_Compiler_Debugger
	CompilerError "No Debugger!!!!!!!!!!"
CompilerEndIf

Procedure ReplaceChar(*String.Character, CharToSearch.c, CharToReplace.c)
	If *String = 0
		ProcedureReturn
	EndIf
	While *String\c
		If *String\c = CharToSearch
			*String\c = CharToReplace
		EndIf
		*String + SizeOf(Character)
	Wend
EndProcedure

OpenConsole()

Define Time.q, Time1.q, Time2.q
Define String.s = ReplaceString(Space(10000), " ", "Abc")

Time = ElapsedMilliseconds()
For I = 1 To 1000
	ReplaceString(String, "A", "a", #PB_String_InPlace)
	ReplaceString(String, "a", "A", #PB_String_InPlace)
Next
Time1 = ElapsedMilliseconds() - Time

Time = ElapsedMilliseconds()
For I = 1 To 1000
	ReplaceChar(@String, Asc("A"), Asc("a"))
	ReplaceChar(@String, Asc("a"), Asc("A"))
Next
Debug String
Time2 = ElapsedMilliseconds() - Time


PrintN("ReplaceString: " + Time1 + " ms")
PrintN("ReplaceChar: " + Time2 + " ms")
Input()

Therefore, as the user moves around the data with PgUp/PgDn/Home/End and so on, we have to convert the control characters into a displayable character, normally 149 which is a dot, so the user can identify the characters that must not be changed or overwritten :

Code: Select all

ReplaceString(Params.s, Chr(60251), Chr(149), #PB_String_InPlace)

You can replace this function and you will gain speed, instead of loosing time.

Code: Select all

ReplaceChar(@Params, 60251, 149)

If used with ReplaceChar(@Test$, 'o', ''), the length needs to be recalculated with PeekS