您现在的位置:首页 >> 基础算法 >> window基础 >> 内容

Pos函数(比FastPos还要快)

时间:2011/9/3 15:35:42 点击:

  核心提示:function QuickPos(const Substr, S: WideString; MatchesIndex: Integer = 1): Integer;function QuickPos...

function QuickPos(const Substr, S: WideString; MatchesIndex: Integer = 1): Integer;
function QuickPosBack(const Substr, S: WideString; MatchesReverseIndex: Integer = 1): Integer;

主要用途是搜索字符串中第n个Substr。
经过测试,这2个函数的速度比直接用Pos+Copy快好几倍(如果字符串够长,可能10几倍)
比Pos+Delete(如JVCL的函数NPos)处理要快(至少2-3倍以上)。

说明1:虽然说速度上的差异是微秒级别的,但是一个快速方便的Pos函数,还是会给编程带来很多方便。
说明2:经过大量测试和FastString作者的承认,FastString的速度还没有DELPHI标准函数快。

========================================================================
// Compares a substring with a string. *for inline use"
// C: 2004-07-05 | M: 2004-07-05
function _InlineCompareText(const Substr, S: WideString; StartIndex: Integer = 1; LenOfSubstr: Integer = -1; LenOfS: Integer = -1): Boolean;
var
  I: Integer;
begin
  if LenOfSubstr = -1 then LenOfSubstr := Length(Substr);
  if LenOfS = -1 then LenOfS := Length(S);
  if LenOfSubstr > LenOfS then
  begin
    Result := False;
    Exit;
  end;
  for I := 1 to LenOfSubstr do
    if Substr[I] <> S[I + StartIndex - 1] then
    begin
      Result := False;
      Exit;
    end;
  Result := True;
end;

// Returns the 1. index of a substring within a string start at a certain index.
// C: 2004-07-05 | M: 2004-07-05 | P: 1.0+
function _PosForward(const Substr, S: WideString; StartIndex: Integer; LenOfSubstr: Integer = -1; LenOfS: Integer = -1): Integer;
var
  I: Integer;
begin
  Result := 0;
  case LenOfSubstr of
    0: Exit;
   -1: LenOfSubstr := Length(Substr);
  end;
  if LenOfS = -1 then LenOfS := Length(S);

  for I := StartIndex to LenOfS do
  begin
    if (S[I] = Substr[1]) and _InlineCompareText(Substr, S, I, LenOfSubstr, LenOfS) then
    begin
      Result := I;
      Exit;
    end;
  end;
end;

// Returns the 1. index of a substring within a string.
// Note: Searching time will increase when MatchesIndex increased.
// C: 2004-04-09 | M: 2004-07-05 | P: 1.0+
function QuickPos(const Substr, S: WideString; MatchesIndex: Integer = 1): Integer;
var
  LenOfS, LenOfSubstr: Integer;
begin
  Result := Pos{Pos}(Substr, S);

  if (MatchesIndex = 1) or (Result = 0) then Exit;
  LenOfS := Length(S);
  LenOfSubstr := Length(Substr);

  while (MatchesIndex > 1) and (Result > 0) do
  begin
    Result := _PosForward{Pos}(Substr, S, Result + 1, LenOfSubstr, LenOfS);  // Tip!! Do not use func.Copy!!
    if Result = 0 then Exit;
    Dec(MatchesIndex);
  end;
end;

// Returns the last index of a substring within a string.
// Todo: Using asm to rewrite this function. The asm-code looks very like func.Pos!
// C: 2004-04-09 | M: 2004-07-03 | P: n/a
function _PosBack(const Substr, S: WideString; StopIndex: Integer = -1; LenOfSubstr: Integer = -1): Integer;
var
  I: Integer;
begin
  Result := 0;
  case LenOfSubstr of
    0: Exit;
   -1: LenOfSubstr := Length(Substr);
  end;
  if StopIndex = -1 then StopIndex := Length(S);
 
  for I := StopIndex - LenOfSubstr + 1 downto 1 do
  begin
    if (S[I] = Substr[1]) and _InlineCompareText(Substr, S, I, LenOfSubstr) then
    begin
      Result := I;
      Exit;
    end;
  end;
end;

// Returns the last index of a substring within a string.
// C: 2004-04-09 | M: 2004-07-03 | P: n/a
function QuickPosBack(const Substr, S: WideString; MatchesReverseIndex: Integer = 1): Integer;
var
  LenOfSubstr: Integer;
begin
  Result := _PosBack{Pos}(Substr, S);

  if (MatchesReverseIndex = 1) or (Result = 0) then Exit;
  LenOfSubstr := Length(Substr);

  while (MatchesReverseIndex > 1) and (Result > 0) do
  begin
    Result := _PosBack{Pos}(Substr, S, Result + LenOfSubstr - 2, LenOfSubstr);
    Dec(MatchesReverseIndex);
  end;
end;  

作者:网络 来源:转载
共有评论 0相关评论
发表我的评论
  • 大名:
  • 内容:
  • 盒子文章(www.2ccc.com) © 2022 版权所有 All Rights Reserved.
  • 沪ICP备05001939号