做了一个 62 进制的简单实现

62 进制需要的字符及次序: 0..9 a..z A..Z; 只实现了 62 进制字符串与正整数的互换.

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unit H62;

interface

uses SysUtils;

function IntToH62(N: UInt64): string; //整数转到 62 进制字符串
function H62ToInt(S: string): UInt64; //62 进制字符串转到整数

implementation

function _C2B(C: Char): Byte; inline;
begin
  Result := 0;
  if CharInSet(C, ['0'..'9']) then Exit(Byte(C) - 48);      //0..9
  if CharInSet(C, ['a'..'z']) then Exit(Byte(C) - 97 + 10); //a..z
  if CharInSet(C, ['A'..'Z']) then Exit(Byte(C) - 65 + 36); //A..Z
end;

function _B2C(B: Byte): Char; inline;
begin
  Result := #0;
  if B <= 9 then Exit(Char(B + 48));                       //0..9
  if (B >= 10) and (B <= 35) then Exit(Char(B - 10 + 97)); //a..z
  if (B >= 36) and (B <= 61) then Exit(Char(B - 36 + 65)); //A..Z
end;

function _Power(B,P: Cardinal): UInt64; inline;
var
  i: Integer;
begin
  Result := B;
  for i := 1 to P-1 do Result := Result * B;
end;

function _C2V(C: Char; N: Byte): UInt64; inline;
begin
  Result := 0;
  if (N = 0) then Exit(_C2B(C));
  if (N > 0) then Result := _C2B(C) * _Power(62, N);
end;

function IntToH62(N: UInt64): string;
var
  C: Char;
begin
  Result := '';
  repeat
    C := _B2C(N mod 62);
    Result := C + Result;
    N := N div 62;
  until (N = 0);
end;

function H62ToInt(S: string): UInt64;
var
  C: Char;
  L,N,I: Cardinal;
begin
  Result := 0;
  L := Length(S);
  if L > 11 then raise Exception.Create('Err: H62ToInt'); //不能多于 11 位
  for I := L downto 1 do
  begin
    C := S[I];
    N := L - I;
    Result := Result + _C2V(C, N);
  end;
end;

end.

---

//测试:
uses H62;

procedure TForm1.FormCreate(Sender: TObject);
var
  n: Cardinal;
  I: UInt64;
  str: string;
begin
  str := IntToH62(MaxInt);  // 2lkCB1
  n := H62ToInt(str);       // 2147483647

  I := 9999999999999999999; // 19 位
  str := IntToH62(I);       // bUI6zOLZTrh

  str := 'ZZZZZZZZZZZ';     // 最大值
  I := H62ToInt(str);       // 15143072536417990655; 比 UInt64 的最大值(18446744073709551615)小一点, 比 Int64 的最大值(9223372036854775807)大一点
end;

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