Fast replace$() function

[mdijkens]

For long time, I've used below replace$() function I created:

Code: (Select All)

Function replace$ (content$, from$, to$)

  content2$ = content$

  flen& = Len(from$): tlen& = Len(to$)

  p& = InStr(content2$, from$)

  If flen& = Len(to$) Then

    Do While p& > 0

      Mid$(content2$, p&, flen&) = to$

      p& = InStr(p& + tlen&, content2$, from$)

    Loop

  Else

    Do While p& > 0

      content2$ = Left$(content2$, p& - 1) + to$ + Mid$(content2$, p& + flen&)

      p& = InStr(p& + tlen&, content2$, from$)

    Loop

  End If

  replace$ = content2$

End Function 

But recently I had to do +3000's replaces times +1500 1MB files.
The total replace-process took over 1100 seconds.

When I realized the amount of memory copying going on when constantly rebuilding the content2$ for each replace, I was wondering if _Mem could be an option.
I ended up with below mreplace$() function which got exactly the same job done in less then 45 seconds!!

Code: (Select All)

 Function mreplace$ (content$, from$, to$)

  Dim mp As Long, pp As Long, m As _MEM

  m = _MemNew(Len(content$) * 2): mp = 0: pp = 1

  flen& = Len(from$): tlen& = Len(to$)

  p& = InStr(content$, from$)

  Do While p& > 0

    _MemPut m, m.OFFSET + mp, Mid$(content$, pp, p& - pp): mp = mp + p& - pp

    _MemPut m, m.OFFSET + mp, to$: mp = mp + tlen&: pp = p& + flen&

    p& = InStr(p& + flen&, content$, from$)

  Loop

  _MemPut m, m.OFFSET + mp, Mid$(content$, pp): mp = mp + Len(Mid$(content$, pp))

  content2$ = String$(mp, 0): _MemGet m, m.OFFSET, content2$: _MemFree m

  mreplace$ = content2$

End Function

Take advantage if you run into this issue