There is a function in the AES algorithm, to multiply a byte by 2 in Galois Field.
This is the function given in a website
private static byte gfmultby02(byte b)
    {
      if (b < 0x80)
        return (byte)(int)(b <<1);
      else
        return (byte)( (int)(b << 1) ^ (int)(0x1b) );
    }
This is the function i wrote.
private static byte MulGF2(byte x)
            {
            if (x < 0x80)
                return (byte)(x << 1);
            else
            {
                return (byte)((x &开发者_开发问答lt;< 1) ^ 0x1b);
            }
}
What i need to know is, given any byte whether this will perform in the same manner. Actually I am worried about the extra cast to int and then again to byte. So far I have tested and it looks fine. Does the extra cast to int and then to byte make a difference in rare cases?
I think in this case the cast to int does nothing, cause the cast is done after the left shift. So let's take a little example:
byte b = 0x1000;
//temp1 == 0x00000000;
int temp1 = (int)(b << 1);
//temp2 == 0x00010000;
int temp2 = ((int)b) << 1);
So as you can see the parentheses have a big impact on the result, but if took the formula from the website right, your code should behave the same.
I think it's correct, but:
The best way to make sure is to simply test it; there are only 256 cases and it shouldn't take many minutes to write the test case.
 
         
                                         
                                         
                                         
                                        ![Interactive visualization of a graph in python [closed]](https://www.devze.com/res/2023/04-10/09/92d32fe8c0d22fb96bd6f6e8b7d1f457.gif) 
                                         
                                         
                                         
                                         加载中,请稍侯......
 加载中,请稍侯......
      
精彩评论