让我开始说我对Delphi一无所知......
我正在尝试将用delphi编写的旧应用程序移植到java中,但是事情不起作用......
我有这个函数对两个字节进行一些二进制操作。这是Delphi中的代码:
function TMainForm.mixthingsup(x, y: byte): word;
var
counter: byte;
answer1, answer2: byte;
begin
answer1 := $9D xor x;
for counter := 1 to 8 do
begin
if (answer1 and $80) = $80 then
answer1 := (answer1 shl 1) xor $26
else
answer1 := (answer1 shl 1);
end;
answer2 := answer1 xor y;
for counter := 1 to 8 do
begin
if ((answer2 and $80) = $80) then
answer2 := ((answer2 shl 1) xor $72)
else
answer2 := (answer2 shl 1);
end;
Result := (answer1 shl 8) or answer2;
end;
这是我的java代码:
public static String mixthingsup(String data)
{
byte[] conv=null;
byte c9d;
byte c80;
byte c26;
byte c72;
byte x,y;
byte res1, res2;
byte res;
conv=hexStringToByteArray(data.substring(0, 2));
x=conv[0];
conv=hexStringToByteArray(data.substring(2, 4));
y=conv[0];
conv=hexStringToByteArray("9d");
c9d=conv[0];
conv=hexStringToByteArray("80");
c80=conv[0];
conv=hexStringToByteArray("26");
c26=conv[0];
conv=hexStringToByteArray("72");
c72=conv[0];
res1=(byte) (c9d ^ x);
for(int i=1; i<9; i++)
{
if((res1 & c80) == c80)
res1=(byte) ((res1 << 1) ^ c26);
else
res1=(byte) (res1 << 1);
}
res2=(byte) (res1 ^ y);
for(int i=1; i<9; i++)
{
if((res2 & c80) == c80)
res2=(byte) ((res2 << 1) ^ c72);
else
res2=(byte) (res2 << 1);
}
res=(byte) ((res1 << 8) | res2);
return Integer.toHexString(res);
}
例如,当delphi函数返回A BA 77的CA BA时,java函数返回FF FF FF BA
有什么想法?有什么帮助吗?
谢谢, 佩德罗
答案 0 :(得分:3)
看看这一行:
res=(byte) ((res1 << 8) | res2);
当您将其转换为byte
时,您将16位值截断为8位,因此丢失res1
。
您应该转换为2字节值short
。
也就是说,在数组中返回两个字节可能更容易。像这样:
public static byte[] MixThingsUp(byte x, byte y)
{
byte answer1 = (byte) (0x9D ^ x);
for (int i=0; i<8; i++)
if ((answer1 & 0x80) == 0x80)
answer1 = (byte) ((answer1 << 1) ^ 0x26);
else
answer1 = (byte) (answer1 << 1);
byte answer2 = (byte) (answer1 ^ y);
for (int i=0; i<8; i++)
if ((answer2 & 0x80) == 0x80)
answer2 = (byte) ((answer2 << 1) ^ 0x72);
else
answer2 = (byte) ((answer2 << 1));
return new byte[] { answer1, answer2 };
}
如果我是你,我会将按位操作和转换与字符串分开。你在问题中的方式混合了两个问题。