位运算 - 算术运算 [英] Bitwise Operations -- Arithmetic Operations
问题描述
能否请您解释一下下面的线,一些很好的例子。
由n个阿左算术移
相当于乘以2 N
(提供的值不
溢出)。
块引用>和
一个正确的算术
由二的补码值的n移位
相当于除以2 N 和
向负无穷大舍入。如果
二进制数被视为那些'
补充,那么同样右移
运算结果除以2 N
和向零舍入。
块引用>解决方案我会解释,我们是比较熟悉的基础发生了什么:10
在基地10,假设你有一个数N = 123。现在,你SHIFT这个数字向左K = 3个位置,以0填充清空位数所以,你得到X = 123000。
注意,X = N * 10 K 。
与基地2的情况是相似的。
例1(基数为10)|实施例2(基数为2)
|
N = 123 | N = 110101(53 10为基数)
K = 3 | K = 2(在基体10)
N'LT;< K = 123000 | N'LT;< K = 11010100(以10为基数212)
|
10 ^ K = 1000 | 2 ^ K = 100(在基体2;在基座10 4)
N * 10 ^ K = 123000 | N * 2 ^ K = 11010100(53 * 4 = 10为基212)
|与右移的情况下仅仅是这个过程的一面镜子,也是在基地10例如类似,如果我在基地10有123456,我转变右3个位置,我得到123这千分之十二万三千四百五十六(整数除法),其中1000 = 10 3 。
Can you please explain the below lines, with some good examples.
A left arithmetic shift by n is equivalent to multiplying by 2n (provided the value does not overflow).
And:
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.
解决方案I will explain what happens in a base that we're more familiar with: 10.
In base 10, let's say you have a number N=123. Now, you "shift" this number to the left k=3 positions, filling the emptied digits with 0. So you get X=123000.
Note that X = N * 10k.
The case with base 2 is analogous.
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) |
The case with right shift is simply a mirror of the process, and is also analogous in base 10. For example, if I have 123456 in base 10, and I "shift" right 3 positions, I get 123. This is 123456 / 1000 (integer division), where 1000 = 103.
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