比 FastPos 还要快 Pos 函数

Moodsky

相信字符串处理中用的最多的就是 Pos 函数了。但是如果要搜索一个字符串中第二次或者第三次出现的子字符串的，就没有现成的 DELPHI 标准函数了。所以我就自己写了一个。同时和网上比较流行的 FastStrings.SmartPos() 和 JVCL.NPos() 做了比较，速度更快，而且兼容 Unicode(WideString/WideChar)。

注：代码可能有人会觉得不太舒服，但作为最常用的字符串函数，这样的优化我觉得还是值得的。


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


代码如下：

// 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;