Best way to replace every third character in a string in delphi - string

Question

What's the most efficient way to replace every third character of the same type in a string?
I have a string like this:
str := 'c1'#9'c2'#9'c3'#9'c4'#9'c5'#9'c6'#9'
I want to replace every third #9 by #13#10, so that i get:
str1 := 'c1'#9'c2'#9'c3'#13#10'c4'#9'c5'#9'c6'#13#10'
I would do this in this way:
i:=0;
newStr:='';
lastPos := Pos(str,#9);
while lastPos > 0 do begin
if i mod 3 = 2 then begin
newStr := newStr + Copy(str,1,lastPos-1) + #13#10;
end else begin
newStr := newStr + Copy(str,1,lastPos);
end;
str := Copy(str,lastPos+1,MaxInt);
i := i + 1;
lastPos := Pos(str,#9);
end;
newStr := Copy(str,1,MaxInt);
But thats a lot of copying. Is there a string manipulation function to do this?

Answer 1

I think the problem as stated doesn't match the code you provided. Is every third character a #9? If so, do you want to change every third appearance of #9 for #13#10?
If so, I would do it this way:
function test(str: string): string;
var
i, c, l: integer;
begin
l := Length(str);
SetLength(Result, l + l div 9);
c := 1;
for i := 1 to l do
begin
if (i mod 9 = 0) and (i > 0) then
begin
Result[c] := #13;
Inc(c);
Result[c] := #10;
end
else
Result[c] := str[i];
Inc(c);
end;
end;
I actually have no idea if this function performs well. But given that the constraints aren't clear, I guess so.
If the position of the #9 character is unknown then this solution won't work at all.
Edit: as David points out, this is not nearly equivalent to the original code posted. This seems to work, but it requires two passes on the original string. The thing is, to know if its more efficient or not we need to know more about the input and context.
function OccurrencesOfChar(const S: string; const C: char): integer;
var
i: integer;
begin
result := 0;
for i := 1 to Length(S) do
if S[i] = C then
inc(result);
end;
function Test(str: string): string;
var
len, n, C, i: integer;
begin
C := 1;
len := Length(str);
n := OccurrencesOfChar(str, #9);
SetLength(result, len + n div 3);
n := 1;
for i := 1 to len do
begin
if str[i] = #9 then
begin
if n mod 3 = 0 then
begin
result[C] := #13;
inc(C);
result[C] := #10;
end
else
result[C] := #9;
Inc(n);
end
else
result[C] := str[i];
inc(C);
end;
end;

Answer 2

I expect this question will be closed, but just for fun, that would be my proposal.
Function Replace(const Instr:String;Re:Char;const ReWith:String):String;
var
i,o,cnt,l:Integer;
begin
cnt:=0;
o:=0;
SetLength(Result,Length(Instr)*Length(ReWith));// just for security
for I := 1 to Length(Instr) do
begin
if Instr[i]=Re then inc(cnt);
if cnt=3 then
begin
for l := 1 to Length(ReWith) do
begin
inc(o);
Result[o] := ReWith[l];
end;
cnt := 0;
end
else
begin
inc(o);
Result[o] := Instr[i];
end;
end;
SetLength(Result,o);
end;
procedure TForm3.Button1Click(Sender: TObject);
begin
Edit2.Text := Replace(Edit1.Text,'A','xxx')
end;

Answer 3

I would probably do something like this (coded in the browser). It only needs one string resize and should have less movement of data around. I exit when I have made the last replacement or if it didn't need any:
procedure ReplaceXChar(var aStringToReplace: string; const aIteration:
Integer; const aChar: Char; const aReplacement: string);
var
replaceCount: Integer;
cntr: Integer;
outputCntr: Integer;
lengthToReplace: Integer;
begin
// Find the number to replace
replaceCount := 0;
for cntr := 1 to Length(aStringToReplace) do
begin
if aStringToReplace[cntr] = aChar then
Inc(replaceCount);
end;
if replaceCount >= aIteration then
begin
// Now replace them
lengthToReplace := Length(aReplacement);
cntr := Length(aStringToReplace);
SetLength(aStringToReplace, cntr +
(replaceCount div aIteration) * (lengthToReplace - 1));
outputCntr := Length(aStringToReplace);
repeat
if aStringToReplace[cntr] = aChar then
begin
if (replaceCount mod aIteration) = 0 then
begin
Dec(outputCntr, lengthToReplace);
Move(aReplacement[1], aStringToReplace[outputCntr+1],
lengthToReplace * SizeOf(Char));
end
else
begin
aStringToReplace[outputCntr] := aStringToReplace[cntr];
Dec(outputCntr);
end;
Dec(replaceCount);
end
else
begin
aStringToReplace[outputCntr] := aStringToReplace[cntr];
Dec(outputCntr);
end;
Dec(cntr);
until replaceCount = 0;
end;
end;
Usage would be like this:
var
myString: String;
begin
myString := 'c1'#9'c2'#9'c3'#9'c4'#9'c5'#9'c6'#9;
ReplaceXChar(myString, 3, #9, #13#10);
ShowMessage(myString);
end;