请问delphi 怎么定义如C语言里面的联合体Union ( 积分: 20 )

int CRC_creat(buf,len)  
 char *buf;
 int len;
{
      long int temp;
      char code[CRC_LEN+1];
      int byte,count,cou;

      temp=(unsigned char )buf[0]*256+(unsigned char )buf[1];

  cou=1;
  while (temp<$10000)
  {
          un.ch=buf[byte+1];

          printf('**** m7:%d',un.bit.m7);

	switch (cou)
	{
		case 1:
			temp=temp*2+un.bit.m7; 
		break;
		case 2:
			temp=temp*2+un.bit.m6; 
		break;
		case 3:
			temp=temp*2+un.bit.m5; 
		break;
		case 4:
			temp=temp*2+un.bit.m4; 
		break;
		case 5:
			temp=temp*2+un.bit.m3; 
		break;
		case 6:
			temp=temp*2+un.bit.m2; 
		break;
		case 7:
			temp=temp*2+un.bit.m1; 
		break;
		case 8:
			temp=temp*2+un.bit.m0; 
		break;
	}


			cou++;
  } /* while <<<<  */

}

function crc_create(var buf:string;len:integer):integer;
var
    temp:LongInt;
    code:string[CRC_LEN];
    byte,count,cou:integer;
    un: TBitUnion;
begin
      un.m7:=1;
      un.m6:=1;
      un.m5:=1;
      un.m4:=1;
      un.m3:=1;
      un.m2:=1;
      un.m1:=1;
      un.m0:=1;

      temp:=(ord(buf[1]))*256+ord(buf[2]);  //注意delphi的string的下标是从1开始
      
      while (temp<$10000) do
         begin 
               un.ch:=buf[byte+1+1]; //注意delphi的string的下表是从1开始的
             
               form1.memo1.lines.add('**** m7: '+inttostr(un.m7));

               case cou of
                    1:temp:=temp*2+un.m7;   
                    2:temp:=temp*2+un.m6;
                    3:temp:=temp*2+un.m5;
                    4:temp:=temp*2+un.m4;
                    5:temp:=temp*2+un.m3;
                    6:temp:=temp*2+un.m2;
                    7:temp:=temp*2+un.m1;
                    8:temp:=temp*2+un.m0;
                end;
            inc(cou);
         end;
end;