Question

If I want to convert 4 bytes into an int, I can do this:

        byte[] b = BitConverter.GetBytes(i1);
        int i2 = BitConverter.ToInt32(b,0);
        int i3 = b[0] | (b[1]<<8) | (b[2]<<16) | (b[3]<<24);

and then i1,i2,i3 will all equal.

but how do I do the same for a uint? This:

        uint u1 = uint.MaxValue-1000;
        byte[] b = BitConverter.GetBytes(u1);
        uint u2 = BitConverter.ToUInt32(b,0);
        uint u3 = (uint)(b[0] | (b[1]<<8) | (b[2]<<16) | (b[3]<<24));

results in a overlflow for large uints.

Answer 1

It would only throw that exception if in a checked context. See: http://msdn.microsoft.com/en-us/library/y3d0kef1(v=vs.80).aspx.

No exception:

uint u1 = uint.MaxValue - 1000;
byte[] b = BitConverter.GetBytes(u1);
uint u2 = BitConverter.ToUInt32(b, 0);
uint u3 = (uint) (b[0] | (b[1] << 8) | (b[2] << 16) | (b[3] << 24));

exception:

checked
{
    uint u1 = uint.MaxValue - 1000;
    byte[] b = BitConverter.GetBytes(u1);
    uint u2 = BitConverter.ToUInt32(b, 0);
    uint u3 = (uint) (b[0] | (b[1] << 8) | (b[2] << 16) | (b[3] << 24));
}

no exception

checked
{
    unchecked
    {
        uint u1 = uint.MaxValue - 1000;

        byte[] b = BitConverter.GetBytes(u1);
        uint u2 = BitConverter.ToUInt32(b, 0);
        uint u3 = (uint) (b[0] | (b[1] << 8) | (b[2] << 16) | (b[3] << 24));

        Console.WriteLine(u1 + " " + u2 + " " + u3);
    }
}

Make sure you’re not compiling with the /checked option.

The exception is thrown by casting from int to uint. Using the shift operator on the bytes (the line with uint u3 = …) implicitly cast them to int. A uint with the MSB on (“1”) is a negative int which is out of range for uint. Using int causes no such exception because there is no explicit cast which might elicit an overflow exception.