Question

Why the and operator acts differently when used:

1. with un unsigned int

2. with a byte array represiting the same value when casted to (uint*)

   unsafe
   {
       fixed (byte* i = new byte[4] { 0x02, 0x03, 0x04, 0xFF })
       {
           uint m = 0x020304FF;
   
           Console.WriteLine("{0:X}",m & 0xFF000000); 
   
           Console.WriteLine("{0:X}",*(uint*)i & 0xFF000000); 
        }
    }
   

result is

     2000000
     ff000000 

Answer 1

Assuming a 32 bits little-endian system (like any Intel processor, in the vast majority of today’s computers), casting the address of the byte-array {02 03 04 FF} to the address of a 32 bit int will result in an int with the value 0xFF040302. Hence the result.

In other words, your assumption that it’s “the same value” is not correct.