Counting number of words

This example searches for strings that are not numbers and have at list 2 characters.
Your definition of "words" can be different and you probably improve this code, because it's not perfect and maybe has some bugs, but just to show you the main idea.

procedure TForm1.Button1Click(Sender: TObject);
var Buf, Point1: pChar;
    I: Word;
begin
  Buf := Memo1.Lines.GetText;
  I:=0;
  Point1:=Buf;
{first we check the spaces}
  while Point1<>nil do
  begin
    if Point1<>nil then
    if not ((Point1[0] in [' ', #13, #10, #0, '0'..'9']) or
      (Point1[1] in [' ', #13, #10, #0, '0'..'9'])) then Inc(I);
    Point1:=StrScan(Point1, ' ');
    if Point1<>nil then
      while Point1[0]=' ' do Inc(Point1);
  end;
{then we check the linefeeds}
  Point1:=Buf;
  Point1:=StrScan(Point1, #10);
  while Point1<>nil do
  begin
    if not ((Point1[1] in [' ', #13, #10, #0,'0'..'9']) or
    (Point1[2] in [' ', #13, #10, #0,'0'..'9'])) then Inc(I);
    Inc(Point1);
    Point1:=StrScan(Point1, #10);
  end;
  StrDispose(Buf);
  Showmessage('number of words is: '+inttostr(i));
end;

Hello y'all.. I just took a peek at the code, and wanted to to comments something... You don't need to StrDispose(Buf) since you never really took memory for that...

Cheers,
Viktor

Hi !

This code nearly did it ! The prob is the improving.
I used ['A'..'Z', 'a'..'z'] instead of [' ', #13, #10, #0, '0'..'9'].
But still some bizarre chars are recognized as words.
I'm not that in that kind og programming.

Any idea ?
--heiko

You can't use ['A'..'Z', 'a'..'z'] instead of [' ', #13, #10, #0, '0'..'9'] because it says NOT before you use [' ', #13, #10, #0, '0'..'9'] which means that this is exacty what you need....try it without changing anything, and then say what you need...

Vikronet, at the end we have to call StrDispose. From Borland's help file:
"...Call GetText to obtain a dynamically allocated character buffer containing all of the strings in the list. Individual strings are separated by a carriage return and line feed. The caller is responsible for freeing the returned value..."<---

Webers, I have showed you the main idea, assuming you know how to continue from that point.
Ronit

Yeah I just read it... You are right ... 10x :)

Cheers,
Viktor

Hi,
there is another way to do this.
You can create a TStringList and then assign Memo.Lines to TStringList.CommaText:

procedure TForm1.Button1Click(Sender: TObject);
var
  sl:TStringList;
  a,count:longint;
begin
  sl:=TStringList.Create;
  count:=0;
  for a:=0 to Memo1.Lines.Count-1 do begin
   sl.Clear;
   sl.CommaText:=Memo1.Lines[a];
   count:=count+sl.Count;
  end;
end;

So, count will be the number of words.
This code recognizes words separated by ' , '   ' ; '  spaces and #10#13 (CR), but if your text is within " " (double quotes), this text will be count as one word.

A.

333, I didn't check your code, but in general, working with pointers should be faster.

Hi !

333, how about this ?
Changes I made aren't the best, but it works.
I fixed the "" problem and now only
"real" words are counted (excluding ?, * and so on).

Hello, y'all up there, especially ronit, thanks
for your answers.
(ronit, I forgot to tell you that I removed the "not"s.)
333, post your comment as answer that I can give
you the points.

take care,
--heiko.


var
  sl: TStringList;
  a, i, count: longint;
  st, ct: string;
  p: Pchar;
begin
  sl := TStringList.Create;
  Count := 0;

  For a := 0 to GetCurMemo.Lines.Count - 1 do begin
   sl.Clear;  //getcurmemo is an Trichedit
   ct:= GetCurmemo.Lines[a];

   while Pos('"', ct) > 0 do   // the "" problem
    Delete(ct, Pos('"', ct), 1);
   sl.CommaText := ct;

    For I := 0 to sl.Count - 1 do begin
      try
       st := sl[I];
       p := Pchar(st);
       if not (p[0] in ['A'..'Z', 'a'..'z', 'Ä', 'ä', 'Ö', 'ö', 'Ü', 'ü', 'ß']) then sl.Delete(I);
      except end; //these signs after 'a'..'z' are german umlauts
    End;
    count := count + sl.Count;
  End;
  ShowMessage(IntToStr(Count));