Replace multiple blanks in string to one blank

var
  S, Temp: string;
begin
  // S contains string to reduce
  Temp := S;
  repeat
    S := Temp;
    Temp := StringReplace(S, '  ', ' ', [rfReplaceAll]);
  until S = Temp;
end;

procedure TForm1.Button1Click(Sender: TObject);
function MyTrim(const S: string): string;
var
  I, L: Integer;
  A: String;
  got: Boolean;
begin
  A := '';
  L := Length(S);
  for i := 1 to L do begin
  If (S[I] <> ' ') then
  begin
  Got := False;
  A := a + s[i];
  end else
   If not Got then
      Begin
         a := a+ ' ';
         Got := True;
      end;
   end;
   Result := A;

end;
begin
Label1.Caption := MyTrim('This     is     a     string');
end;

my version:

function RemoveRedundantSpaces(const S: string): string;
var
  sList: TStringList;
  i: Integer;
begin
  sList := TStringList.Create;
  Result := '';
  with sList do
  try
    Text := StringReplace(S, ' ', #13, [rfReplaceAll]);
    for i := 0 to Count - 1 do
      if Strings[i] <> '' then
        Result := Result + Strings[i] + ' ';
    Result := Trim(Result);
  finally
    Free;
  end;
end;

to use it:

Label1.Caption := RemoveRedundantSpace('This       is        a           string');

note:

shortest note doesn't mean most efficient ;-)

PS: Of course, neither was I saying that mine was the most efficient :-p

This routine will strip out all of the multiple spaces in the string in only one pass (unlike the shorter but less efficient "StringReplace" example).

function StripMultiSpaces( const AString:string ):string;
var
  Index:integer;
  OnSpace:boolean;
begin
  Result := '';
  OnSpace := False;
  for Index := 1 to Length(AString) do begin
    if (AString[Index] = ' ') then begin
      if not OnSpace then Result := Result + AString[Index];
    end else begin
      Result := Result + AString[Index];
    end; {if}
    OnSpace := (AString[Index] = ' ');
  end; {for}
end;

Oops, Ferruccio68's algorithm is pretty much the same as the one I just posted (his uses more local variables though).

Hi,

My version:

function RemoveRedundantSpaces(const S: string): string;
var
  Len: integer;
  i,j: integer;
  Former: char;
begin
  result := S;
  Len := Length(S);
  Former := S[1];
  j := 2;
  for i := 2 to Len do begin
    if (S[i] = ' ') then begin
      if (not (Former = ' ')) then begin
        result[j] := S[i];
        Inc(j);
      end;
    end else begin
      result[j] := S[i];
      Inc(j);
    end;
    Former := S[i];
  end;
  SetLength(result, j);
end;

Regards, Geo

It is quite important that the function is fast too ofcourse.
So a onepass through the string is very much appreciated.

My solution is ten times faster than the accepted one :-(

ehhh Geo, that's the life :)))

F68 ;-)

...the accepted answer also tramples off the end of the string if you give it a single character. That said, it isn't really an issue as it only tramples onto the #0 terminating the string.

F68, your function is the second best of the six. Testing agains huge strings my function becomes more than 50 times faster compared to the accepted one.

Best regards,

I see geo, and i agree with you, but you and me know how to test and choose the best one. Maybe he did like the accepted one just for the 'less code' without give a try to every posted code...

Every man is different from each other man, that's why i said 'that's the life' :)))

F68 ;-)

Sure, F68. I put effords on testing because Delphiwizard said:
>It is quite important that the function is fast too ofcourse.
and posted the results of the tests for the eventual future readers ;-)

The reason that I choose this one is probably because it worked fine and it has not to many lines of code.
If the other code of geobul is faster, that I don't know.
If I was smart enough to think of the proposed solutions and be able to determine which is faster then it probably would have chosen different. I try to choose better next time.
Thanks anyway for your input.