This question is NOT a duplicate of this question.
I came across a situation where I might have had to left-shift a (positive) number by a negative value, i.e., 8 << -1. In that case, I would expect the result to be 4, but I'd never done this before. So I made up a little test program to verify my hypothesis:
for (int i = -8; i <= 4; i++)
Console.WriteLine("i = {0}, 8 << {0} = {1}", i, 8 << i);
which to my shock and surprise gave me the following output:
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
Can anyone explain this behaviour?
Here's a little bonus. I changed the left-shift to a right-shift, and got this output:
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
In C-like languages << -1
doesn't translate to >> 1
. Instead the least-significant 5 bits of the shift are taken and the rest ignored, so in this case the two's complement -1
translates to << 31
.
You'll get the same results from eg. JavaScript javascript:alert(8<<-8)
.
You can't shift by a negative value. You also can't shift by a large positive number.
From the C# spec (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.
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