求一个base64算法(100分)

zhuangqr · 2006-11-18

求一个base64算法  急~~!

jxgxy · 2006-11-18

//---------------------------------------------------------------------------- //base64字符串加密函数 function Base64Encode(const s: string): string; var  i,c1,c2,c3: Integer;  m,n: Integer; const  Base64: string = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'; begin  Result := '';  m:=1;  n:=0;  for i := 1 to (Length(s) div 3) do  begin    c1 := Ord(s[m]);    c2 := Ord(s[m+1]);    c3 := Ord(s[m+2]);    m:=m+3;    Result := Result+base64[(c1 shr 2)and $3F+1];    Result := Result+base64[((c1 shl 4)and $30) or ((c2 shr 4)and $0F)+1];    Result := Result+base64[((c2 shl 2)and $3C) or ((c3 shr 6)and $03)+1];    Result := Result+base64[c3 and $3F+1];    n:=n+4;    if(n = 76)then    begin       n:=0;       Result := Result+#13#10;    end;  end;  if (Length(s) mod 3)=1 then  begin    c1 := Ord(s[m]);    Result := Result+base64[(c1 shr 2)and $3F+1];    Result := Result+base64[(c1 shl 4)and $30+1];    Result := Result+'=';    Result := Result+'=';  end;  if (Length(s) mod 3)=2 then  begin    c1 := Ord(s[m]);    c2 := Ord(s[m+1]);    Result := Result+ base64[(c1 shr 2)and $3F+1];    Result := Result+ base64[((c1 shl 4)and $30) or ((c2 shr 4)and $0F)+1];    Result := Result+base64[(c2 shl 2)and $3C+1];    Result := Result+ '=';  end; end; //---------------------------------------------------------------------------- //base64字符串解密函数 function Base64Decode(const s: string): string; var  i,m,n: Integer;  c1,c2,c3,c4: Integer; const  Base64: string = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'; begin  Result := '';  n:=1;  m:=Length(s);  if s[m]='='then m:=m-1;  if s[m]='='then m:=m-1;  for i:=1 to m div 4 do  begin    c1:=Pos(s[n],Base64)-1;    c2:=Pos(s[n+1],Base64)-1;    c3:=Pos(s[n+2],Base64)-1;    c4:=Pos(s[n+3],Base64)-1;    n:=n+4;    Result:=Result+Chr(((c1 shl 2)and $FC)or((c2 shr 4)and $3));    Result:=Result+Chr(((c2 shl 4)and $F0)or((c3 shr 2)and $0F));    Result:=Result+Chr(((c3 shl 6)and $C0)or c4);  end;  if m mod 4=2 then  begin    c1:=Pos(s[n],Base64)-1;    c2:=Pos(s[n+1],Base64)-1;    Result:=Result+Chr(((c1 shl 2)and $FC)or((c2 shr 4)and $3));  end;  if m mod 4=3 then  begin    c1:=Pos(s[n],Base64)-1;    c2:=Pos(s[n+1],Base64)-1;    c3:=Pos(s[n+2],Base64)-1;    Result:=Result+Chr(((c1 shl 2)and $FC)or((c2 shr 4)and $3));    Result:=Result+Chr(((c2 shl 4)and $F0)or((c3 shr 2)and $0F));  end; end; 分全部给我吧,我没分了.

zhuangqr · 2006-11-19

符合要求

求一个base64算法(100分)

zhuangqr

Unregistered / Unconfirmed

jxgxy

Unregistered / Unconfirmed

zhuangqr

Unregistered / Unconfirmed

Similar threads