看代码如下:
var
  s: Pointer;
  pd: PWORD;
  size, leng, i: Integer;
  str,file1: string;  
begin
...
 DecompressBuf(s, Size, 0, Pointer(pd), leng); str:=string(pd);
//执行这句话,str的结束如下:
//#8#8#$A#8#9#8#6#8#2#8#6#8#2#8#6#8#6#8#5#8#5#8#5#8#5#8#4#8#0#8#7#8#9#8#9#8#5#8#6#8#1#8#4#8#3#8#//6#8#5#8#6#8#5#8#5#8#5#8#4#8#5#8#5#8#4#8#6#8#6#8#$A#8#7#8#7#8#8#8#6#8#6#8#6#8#3#8#1#8#1#8#3#8#3//#8#2#8#3#8#6#8#7#8#9#8#7#8#8#8#8#8#$A#8#9#8#8#8#5#8'??'#5#8#4#8#5#8#5#8#9#8#7#8#4#8#0#8#3#8#6#//8#2#8#5#8#9#8#7#8#  //执行以下就会得到2056,2058,......
  for i:=0 to leng div 2  do begin
    str := str + IntToStr(pd^) +',';
    inc(pd);
  end;end;问题: 2056,2058,...... 这些数据是哪来的呢..
本人对Delphi不熟, 请大师指点.  
本人用的是.Net的, 在.Net方式可以获取到
//#8#8#$A#8#9#8#6#8#2#8#6#8#2#8#6#8#6#8#5#8#5#8#5#8#5#8#4#8#0#8#7#8#9#8#9#8#5#8#6#8#1#8#4#8#3#8#//6#8#5#8#6#8#5#8#5#8#5#8#4#8#5#8#5#8#4#8#6#8#6#8#$A#8#7#8#7#8#8#8#6#8#6#8#6#8#3#8#1#8#1#8#3#8#3//#8#2#8#3#8#6#8#7#8#9#8#7#8#8#8#8#8#$A#8#9#8#8#8#5#8'??'#5#8#4#8#5#8#5#8#9#8#7#8#4#8#0#8#3#8#6#//8#2#8#5#8#9#8#7#8#现在要怎么把这些数据转成2056,2058,...... 呢...
DecompressBuf