Question

请解释如下内容,并举一些很好的例子。

A left arithmetic shift by n is equivalent to multiplying by 2ⁿ (provided the value does not overflow).

而且:

A right arithmetic shift by n of a two s complement value is equivalent to dividing by 2ⁿ and rounding toward negative infinity. If the binary number is treated as ones complement, then the same right-shift operation results in division by 2ⁿ and rounding toward zero.

Answer 1

我将解释我们更熟悉的以下基础发生的情况:10。

在10号基数中,请说你有编号N=123。现在,你把这个人数“临时”到左克=3个职位,用零填充empt的位数。因此,你获得X=123000。

注:X = N * 10 ^k。

基准2的情况类似。

 Example 1 (base 10)   |  Example 2 (base 2)
                       |
 N        =    123     |  N       =   110101 (53 in base 10)
 k        =      3     |  k       =        2 (in base 10)
 N << k   = 123000     |  N << k  = 11010100 (212 in base 10)
                       |
 10^k     =   1000     |  2^k     =      100 (in base 2; 4 in base 10)
 N * 10^k = 123000     |  N * 2^k = 11010100 (53 * 4 = 212 in base 10)
                       |

正确转变的情况只是进程的一种反映,也类似于第10条。例如,如果我有123456个基数10和1个“临时”右3职位,我就有了123个。这是123456 /1000(合并司),其中1000 = 10³。

Answer 2

http://en.wikipedia.org/wiki/Arithmetic_second

Answer 3

很容易树立你自己的榜样。

考虑在双文本中添加以下五条:101。左轮转,接下10<>10,为10。页: 1

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