Why should I explicitly surround with “unchecked”?

Question

Is there anyone able to explain me this strange behavior?

    int i = 0x1234;
    byte b1 = (byte)i;
    byte b2 = (byte)0x1234;         //error: const value         


        

Answer 1

You shouldn't surround this with unchecked. Unchecked allows assignment of dangerous value types to a type, which may cause overflows.

byte b1 = (byte)i; will cause an overflow or cast exception at runtime.

byte b2 = (byte)0x1234; is invalid because you can't store values larger than 0xFF in a byte.

byte b3 = unchecked((byte)0x1234); will place either 0x34 or 0x12 (depending on the CLR implementation) into b3, and the other byte will overflow.

byte b4 = checked((byte)i); is the same as byte b1 = (byte)i;

byte b5 = (byte)(int)0x1234; will cast 0x1234 to an int, and then try to cast it to byte. Again, you can't convert 0x1234 to a byte because it's too large.