为什么我得到一个奇怪的结果位移负值？

Question

此问题不是this question的重复。

我遇到了一种情况，我可能不得不左移一个负值，即8 < < -1。在那种情况下，我希望结果是4，但我从来没有这样做过。所以我做了一个小测试程序来验证我的假设：

for (int i = -8; i <= 4; i++) 
    Console.WriteLine("i = {0}, 8 << {0} = {1}", i, 8 << i);

这让我震惊和惊喜给我下面的输出：

i = -8, 8 << -8 = 134217728 
i = -7, 8 << -7 = 268435456 
i = -6, 8 << -6 = 536870912 
i = -5, 8 << -5 = 1073741824 
i = -4, 8 << -4 = -2147483648 
i = -3, 8 << -3 = 0 
i = -2, 8 << -2 = 0 
i = -1, 8 << -1 = 0 
i = 0, 8 << 0 = 8 
i = 1, 8 << 1 = 16 
i = 2, 8 << 2 = 32 
i = 3, 8 << 3 = 64 
i = 4, 8 << 4 = 128

任何人都可以解释这种现象？

这里有一点奖励。我改变了左移到右移，并得到这个输出：

i = -8, 8 >> -8 = 0 
i = -7, 8 >> -7 = 0 
i = -6, 8 >> -6 = 0 
i = -5, 8 >> -5 = 0 
i = -4, 8 >> -4 = 0 
i = -3, 8 >> -3 = 0 
i = -2, 8 >> -2 = 0 
i = -1, 8 >> -1 = 0 
i = 0, 8 >> 0 = 8 
i = 1, 8 >> 1 = 4 
i = 2, 8 >> 2 = 2 
i = 3, 8 >> 3 = 1 
i = 4, 8 >> 4 = 0

Answer 1

你不能移动一个负值。你也不能移动一个大的正数。

从C＃规范（http://msdn.microsoft.com/en-us/library/a1sway8w.aspx）：

If first operand is an int or uint (32-bit quantity), 
the shift count is given by the low-order five bits of second operand. 

... 


The high-order bits of first operand are discarded and the low-order 
empty bits are zero-filled. Shift operations never cause overflows. 

Answer 2

在<< -1不转化为>> 1类似C语言。取而代之的是移位的最低有效位5位被忽略，所以在这种情况下，二进制补码-1转换为<< 31。

你会从例如。 JavaScript javascript:alert(8<<-8)。