Try & Error

우선, 모델이 되는 숫자 하나를 골라볼까요? ^^

uint un = 2320987651;
int n = 2020987651;
OutputBitText(un + "[unsigned] ==> \t", un);
OutputBitText(n + "[signed]   ==> \t", n);

private static void OutputBitText(string text, uint n)
{
    Console.WriteLine(text + " " + Convert.ToString(n, 2).PadLeft(32, '0'));
}

private static void OutputBitText(string text, int n)
{
    Console.WriteLine(text + " " + Convert.ToString(n, 2).PadLeft(32, '0'));
}

// 출력 결과
2320987651[unsigned] ==>         10001010010101110111011000000011, 2320987651
-1753885949[signed]   ==>        10010111011101011101001100000011, -1753885949

2320987651[unsigned] ==>         10001010010101110111011000000011, 2320987651
[unsigned] >> 4 ==>              00001000101001010111011101100000, 145061728
[unsigned] << 4 ==>              10100101011101110110000000110000, 2776064048

-1753885949[signed]   ==>        10010111011101011101001100000011, -1753885949
[signed]   >> 4 ==>              11111001011101110101110100110000, -109617872
[signed]   << 4 ==>              01110111010111010011000000110000, 2002595888

unsigned 의 경우, 밀려난 비트 들에 대한 처리를 하지 않고 최상위 비트에 대한 처리도 없음
signed 의 경우, 
    우측 shift 연산자는 최상위 부호 비트를 유지하면서 밀려난 비트들에 대한 처리를 하지 않고,
    좌측 shift 연산자는 부호 비트를 지키지 않고, 역시 밀려난 비트들에 대한 처리를 하지 않는다.

C# bitwise rotate left and rotate right
; http://stackoverflow.com/questions/812022/c-sharp-bitwise-rotate-left-and-rotate-right

static uint UnsignedLeftShift(uint number, int shift)
{
    return (uint)((number << shift) | (number >> (32 - shift)));
}

static uint UnsignedRightShift(uint number, int shift)
{
    return (uint)((number >> shift) | (number << (32 - shift)));
}

OutputBitText("[rotation] >> 4 ==> \t\t", UnsignedRightShift(un, 4));
OutputBitText("[rotation] << 4 ==> \t\t", UnsignedLeftShift(un, 4));

OutputBitText("[rotation] >> 4 ==> \t\t", UnsignedRightShift((uint)n, 4));
OutputBitText("[rotation] << 4 ==> \t\t", UnsignedLeftShift((uint)n, 4));

// 출력 결과
2320987651[unsigned] ==>         10001010010101110111011000000011, 2320987651
[rotation] >> 4 ==>              00111000101001010111011101100000, 950368096
[rotation] << 4 ==>              10100101011101110110000000111000, 2776064056

-1753885949[signed]   ==>        10010111011101011101001100000011, -1753885949
[rotation] >> 4 ==>              00111001011101110101110100110000, 964123952
[rotation] << 4 ==>              01110111010111010011000000111001, 2002595897

==== Java ====
   n       == 320987651  == 00010011001000011110001000000011
  -n       == -320987651 == 11101100110111100001110111111101

   n >>> 3 == 40123456   == 00000010011001000011110001000000 
  -n >>> 3 == 496747455  == 00011101100110111100001110111111

bitwise unsigned right shift >>>
; http://social.msdn.microsoft.com/Forums/en-US/csharpgeneral/thread/6c892d10-75ff-4f4e-a555-3017f72ec170/

static int TripleRightShift(int number, int shift)
{
    return (int)((uint)number >> shift);
}

// 출력 결과
320987651  >>> 3 ==  00000010011001000011110001000000, 40123456
-320987651 >>> 3 ==  00011101100110111100001110111111, 496747455

C# - Right operand 가 음수인 Shift 연산 결과
; http://www.sysnet.pe.kr/2/0/1008

[출처] C# 의 Shift 비트 연산 정리|작성자 techshare