CRC-6 GSM查表计算结果错误，求Pascal代码修正方案

我尝试用查表法计算CRC-6 GSM，计算字符串'123456789'时得到结果$16，但正确结果应为$13。

以下是我的Pascal实现代码：

var Table: array[0..255] of Byte = (
$3F, $10, $0E, $21, $32, $1D, $03, $2C, 
$25, $0A, $14, $3B, $28, $07, $19, $36, 
$0B, $24, $3A, $15, $06, $29, $37, $18, 
$11, $3E, $20, $0F, $1C, $33, $2D, $02, 
$38, $17, $09, $26, $35, $1A, $04, $2B, 
$22, $0D, $13, $3C, $2F, $00, $1E, $31, 
$0C, $23, $3D, $12, $01, $2E, $30, $1F, 
$16, $39, $27, $08, $1B, $34, $2A, $05, 
$31, $1E, $00, $2F, $3C, $13, $0D, $22, 
$2B, $04, $1A, $35, $26, $09, $17, $38, 
$05, $2A, $34, $1B, $08, $27, $39, $16, 
$1F, $30, $2E, $01, $12, $3D, $23, $0C, 
$36, $19, $07, $28, $3B, $14, $0A, $25, 
$2C, $03, $1D, $32, $21, $0E, $10, $3F, 
$02, $2D, $33, $1C, $0F, $20, $3E, $11, 
$18, $37, $29, $06, $15, $3A, $24, $0B, 
$23, $0C, $12, $3D, $2E, $01, $1F, $30, 
$39, $16, $08, $27, $34, $1B, $05, $2A, 
$17, $38, $26, $09, $1A, $35, $2B, $04, 
$0D, $22, $3C, $13, $00, $2F, $31, $1E, 
$24, $0B, $15, $3A, $29, $06, $18, $37, 
$3E, $11, $0F, $20, $33, $1C, $02, $2D, 
$10, $3F, $21, $0E, $1D, $32, $2C, $03, 
$0A, $25, $3B, $14, $07, $28, $36, $19, 
$2D, $02, $1C, $33, $20, $0F, $11, $3E, 
$37, $18, $06, $29, $3A, $15, $0B, $24, 
$19, $36, $28, $07, $14, $3B, $25, $0A, 
$03, $2C, $32, $1D, $0E, $21, $3F, $10, 
$2A, $05, $1B, $34, $27, $08, $16, $39, 
$30, $1F, $01, $2E, $3D, $12, $0C, $23, 
$1E, $31, $2F, $00, $13, $3C, $22, $0D, 
$04, $2B, $35, $1A, $09, $26, $38, $17
);

function GSM_Update(Msg: PByte; Length: Integer): Byte;
var i: Integer;
    H: Byte;
begin
  H :=  $00;

  for i:=0 to Length-1 do begin

    H := Table[H xor Msg^];  // 我怀疑这一行有问题，但不知道怎么改
    Inc(Msg);
  end;

  Result := H xor $3F;
end;

调用方式：

var Msg: AnsiString;
...
GSM_Update(@Msg[1], Length(Msg));

或者

var Msg: array of AnsiChar;
Len: Integer;
...
SetLength(Msg, Len); 
GSM_Update(@Msg[0], Len);

问题分析与修正 #

CRC-6 GSM的核心算法规则是：初始CRC值为$3F，输入字节取高6位参与运算，最终结果无需额外异或。你的代码存在三个关键错误：

1. 初始值错误：代码中H初始化为$00，但标准CRC-6 GSM要求初始值为$3F。
2. 字节处理逻辑错误：直接用完整字节和CRC值异或，忽略了CRC-6仅使用6位的特性，需将输入字节右移2位（取高6位）后再与CRC值异或。
3. 冗余的最终异或：结尾的H xor $3F是多余操作，标准算法不需要这一步。

修正后的函数代码：

function GSM_Update(Msg: PByte; Length: Integer): Byte;
var i: Integer;
    H: Byte;
begin
  H := $3F; // 修正初始值

  for i := 0 to Length-1 do begin
    // 取输入字节高6位，与当前CRC值异或后查表
    H := Table[(H xor (Msg^ shr 2)) and $3F];
    Inc(Msg);
  end;

  Result := H; // 移除多余异或
end;

如果你的CRC表是正确的，使用修正后的代码计算'123456789'将得到正确结果$13。

内容的提问来源于stack exchange，提问作者Tom