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;
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