Quick padding of a string in Delphi

Question

I was trying to speed up a certain routine in an application, and my profiler, AQTime, identified one method in particular as a bottleneck. The method has been with us for years, and is part of a "misc"-unit:

function cwLeftPad(aString:string; aCharCount:integer; aChar:char): string;
var
  i,vLength:integer;
begin
  Result := aString;
  vLength := Length(aString);
  for I := (vLength + 1) to aCharCount do    
    Result := aChar + Result;
end;

In the part of the program that I'm optimizing at the moment the method was called ~35k times, and it took a stunning 56% of the execution time!

It's easy to see that it's a horrible way to left-pad a string, so I replaced it with

function cwLeftPad(const aString:string; aCharCount:integer; aChar:char): string; 
begin
  Result := StringOfChar(aChar, aCharCount-length(aString))+aString;
end;

which gave a significant boost. Total running time went from 10,2 sec to 5,4 sec. Awesome! But, cwLeftPad still accounts for about 13% of the total running time. Is there an easy way to optimize this method further?

Answer 1

Your new function involves three strings, the input, the result from StringOfChar, and the function result. One of them gets destroyed when your function returns. You could do it in two, with nothing getting destroyed or re-allocated.

1. Allocate a string of the total required length.
2. Fill the first portion of it with your padding character.
3. Fill the rest of it with the input string.

Here's an example:

function cwLeftPad(const aString: AnsiString; aCharCount: Integer; aChar: AnsiChar): AnsiString;
var
  PadCount: Integer;
begin
  PadCount := ACharCount - Length(AString);
  if PadCount > 0 then begin
    SetLength(Result, ACharCount);
    FillChar(Result[1], PadCount, AChar);
    Move(AString[1], Result[PadCount + 1], Length(AString));
  end else
    Result := AString;
end;

I don't know whether Delphi 2009 and later provide a double-byte Char-based equivalent of FillChar, and if they do, I don't know what it's called, so I have changed the signature of the function to explicitly use AnsiString. If you need WideString or UnicodeString, you'll have to find the FillChar replacement that handles two-byte characters. (FillChar has a confusing name as of Delphi 2009 since it doesn't handle full-sized Char values.)

Another thing to consider is whether you really need to call that function so often in the first place. The fastest code is the code that never runs.

Answer 2

Another thought - if this is Delphi 2009 or 2010, disable "String format checking" in Project, Options, Delphi Compiler, Compiling, Code Generation.

Answer 3

StringOfChar is very fast and I doubt you can improve this code a lot. Still, try this one, maybe it's faster:

function cwLeftPad(aString:string; aCharCount:integer; aChar:char): string;
var
  i,vLength:integer;
  origSize: integer;
begin
  Result := aString;
  origSize := Length(Result);
  if aCharCount <= origSize then
    Exit;
  SetLength(Result, aCharCount);
  Move(Result[1], Result[aCharCount-origSize+1], origSize * SizeOf(char));
  for i := 1 to aCharCount - origSize do
    Result[i] := aChar;
end;

EDIT: I did some testing and my function is slower than your improved cwLeftPad. But I found something else - there's no way your CPU needs 5 seconds to execute 35k cwLeftPad functions except if you're running on PC XT or formatting gigabyte strings.

I tested with this simple code

for i := 1 to 35000 do begin
  a := 'abcd1234';
  b := cwLeftPad(a, 73, '.');
end;

and I got 255 milliseconds for your original cwLeftPad, 8 milliseconds for your improved cwLeftPad and 16 milliseconds for my version.

Answer 4

You call StringOfChar every time now. Of course this method checks if it has something to do and jumps out if length is small enough, but maybe the call to StringOfChar is time consuming, because internally it does another call before jumping out.

So my first idea would be to jump out by myself if there is nothing to do:

function cwLeftPad(const aString: string; aCharCount: Integer; aChar: Char;): string;
var
  l_restLength: Integer;
begin
  Result  := aString;
  l_restLength := aCharCount - Length(aString);
  if (l_restLength < 1) then
    exit;

  Result := StringOfChar(aChar, l_restLength) + aString;
end;

Answer 5

You can speed up this routine even more by using lookup array.

Of course it depends on your requirements. If you don't mind wasting some memory... I guess that the function is called 35 k times but it has not 35000 different padding lengths and many different chars.

So if you know (or you are able to estimate in some quick way) the range of paddings and the padding chars you could build an two-dimensional array which include those parameters. For the sake of simplicity I assume that you have 10 different padding lengths and you are padding with one character - '.', so in example it will be one-dimensional array.

You implement it like this:

type
  TPaddingArray = array of String;

var
  PaddingArray: TPaddingArray;
  TestString: String;

function cwLeftPad4(const aString:string; const aCharCount:integer; const aChar:char; var anArray: TPaddingArray ): string;
begin
  Result := anArray[aCharCount-length(aString)] + aString;
end;

begin
  //fill up the array
  SetLength(StrArray, 10);
  PaddingArray[0] := '';
  PaddingArray[1] := '.';
  PaddingArray[2] := '..';
  PaddingArray[3] := '...';
  PaddingArray[4] := '....';
  PaddingArray[5] := '.....';
  PaddingArray[6] := '......';
  PaddingArray[7] := '.......';
  PaddingArray[8] := '........';
  PaddingArray[9] := '.........';

  //and you call it..
  TestString := cwLeftPad4('Some string', 20, '.', PaddingArray);
end;

Here are benchmark results:

Time1 - oryginal cwLeftPad          : 27,0043604142394 ms.
Time2 - your modyfication cwLeftPad : 9,25971967336897 ms.
Time3 - Rob Kennedy's version       : 7,64538131122457 ms.
Time4 - cwLeftPad4                  : 6,6417059620664 ms.

Updated benchmarks:

Time1 - oryginal cwLeftPad          : 26,8360194218451 ms.
Time2 - your modyfication cwLeftPad : 9,69653117046119 ms.
Time3 - Rob Kennedy's version       : 7,71149259179622 ms.
Time4 - cwLeftPad4                  : 6,58248533610693 ms.
Time5 - JosephStyons's version      : 8,76641780969192 ms.

The question is: is it worth the hassle?;-)

Answer 6

It's possible that it may be quicker to use StringOfChar to allocate an entirely new string the length of string and padding and then use move to copy the existing text over the back of it.
My thinking is that you create two new strings above (one with FillChar and one with the plus). This requires two memory allocates and constructions of the string pseudo-object. This will be slow. It may be quicker to waste a few CPU cycles doing some redundant filling to avoid the extra memory operations.
It may be even quicker if you allocated the memory space then did a FillChar and a Move, but the extra fn call may slow that down.
These things are often trial-and-error!

Answer 7

You can get dramatically better performance if you pre-allocate the string.

function cwLeftPadMine
{$IFDEF VER210}  //delphi 2010
(aString: ansistring; aCharCount: integer; aChar: ansichar): ansistring;
{$ELSE}
(aString: string; aCharCount: integer; aChar: char): string;
{$ENDIF}
var
  i,n,padCount: integer;
begin
  padCount := aCharCount - Length(aString);

  if padCount > 0 then begin
    //go ahead and set Result to what it's final length will be
    SetLength(Result,aCharCount);
    //pre-fill with our pad character
    FillChar(Result[1],aCharCount,aChar);

    //begin after the padding should stop, and restore the original to the end
    n := 1;
    for i := padCount+1 to aCharCount do begin
      Result[i] := aString[n];
    end;
  end
  else begin
    Result := aString;
  end;
end;

And here is a template that is useful for doing comparisons:

procedure TForm1.btnPadTestClick(Sender: TObject);
const
  c_EvalCount = 5000;  //how many times will we run the test?
  c_PadHowMany = 1000;  //how many characters will we pad
  c_PadChar = 'x';  //what is our pad character?
var
  startTime, endTime, freq: Int64;
  i: integer;
  secondsTaken: double;
  padIt: string;
begin
  //store the input locally
  padIt := edtPadInput.Text;

  //display the results on the screen for reference
  //(but we aren't testing performance, yet)
  edtPadOutput.Text := cwLeftPad(padIt,c_PadHowMany,c_PadChar);

  //get the frequency interval of the OS timer    
  QueryPerformanceFrequency(freq);

  //get the time before our test begins
  QueryPerformanceCounter(startTime);

  //repeat the test as many times as we like
  for i := 0 to c_EvalCount - 1 do begin
    cwLeftPad(padIt,c_PadHowMany,c_PadChar);
  end;

  //get the time after the tests are done
  QueryPerformanceCounter(endTime);

  //translate internal time to # of seconds and display evals / second
  secondsTaken := (endTime - startTime) / freq;
  if secondsTaken > 0 then begin
    ShowMessage('Eval/sec = ' + FormatFloat('#,###,###,###,##0',
      (c_EvalCount/secondsTaken)));
  end
  else begin
    ShowMessage('No time has passed');
  end;
end;

Using that benchmark template, I get the following results:

The original: 5,000 / second
Your first revision: 2.4 million / second
My version: 3.9 million / second
Rob Kennedy's version: 3.9 million / second

Answer 8

This is my solution. I use StringOfChar instead of FillChar because it can handle unicode strings/characters:

function PadLeft(const Str: string; Ch: Char; Count: Integer): string;
begin
  if Length(Str) < Count then
  begin
    Result := StringOfChar(Ch, Count);
    Move(Str[1], Result[Count - Length(Str) + 1], Length(Str) * SizeOf(Char));
  end
  else Result := Str;
end;

function PadRight(const Str: string; Ch: Char; Count: Integer): string;
begin
  if Length(Str) < Count then
  begin
    Result := StringOfChar(Ch, Count);
    Move(Str[1], Result[1], Length(Str) * SizeOf(Char));
  end
  else Result := Str;
end;

Answer 9

It's a bit faster if you store the length of the original string in a variable:

function PadLeft(const Str: string; Ch: Char; Count: Integer): string;
var
  Len: Integer;
begin
  Len := Length(Str);
  if Len < Count then
  begin
    Result := StringOfChar(Ch, Count);
    Move(Str[1], Result[Count - Len + 1], Len * SizeOf(Char));
  end
  else Result := Str;
end;

function PadRight(const Str: string; Ch: Char; Count: Integer): string;
var
  Len: Integer;
begin
  Len := Length(Str);
  if Len < Count then
  begin
    Result := StringOfChar(Ch, Count);
    Move(Str[1], Result[1], Len * SizeOf(Char));
  end
  else Result := Str;
end;