blocks|key|145034|text|位移位运算符实际上并不除以2。相反，它将数字的位数向右移动右手边给定的位置数。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|145035|00101100+=+44
00010110+=+44+>>+1+=+22|code-block|syntax|javascript|145036|请注意，第二行中的位与上面的行相同，只是向右移动了一个位置。现在看第二个例子：|145037|00001011+=+11
00000101+=+11+>>+1+=+5|145038|这与之前的操作完全相同。然而，5的结果是因为最后一位向右移动并消失，从而产生结果5。由于这种行为，右移位运算符通常等同于除以2，然后丢弃任何余数或小数部分。|145039|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The bit shift operator doesn't actually divide by 2. Instead, it moves the bits of the number to the right by the number of positions given on the right hand side. For example:

<pre><code>00101100 = 44
00010110 = 44 &gt;&gt; 1 = 22
</code></pre>

Notice how the bits in the second line are the same as the line above, merely
shifted one place to the right. Now look at the second example:

<pre><code>00001011 = 11
00000101 = 11 &gt;&gt; 1 = 5
</code></pre>

This is exactly the same operation as before. However, the result of 5 is due to the fact that the last bit is shifted to the right and disappears, creating the result 5. Because of this behavior, the right-shift operator will generally be equivalent to dividing by two and then throwing away any remainder or decimal portion.

blocks|key|1252972|text|二进制中的11是1011|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1252973|11+>>+1+|code-block|syntax|javascript|1252974|意味着你将你的二进制表示向右移动一步。|1252975|1011+>>+1+=+101|1252976|然后你有101的二进制，它是1*1+%2B+0*2+%2B+1*4+=+5。|1252977|如果你做了11+>>+2，你会得到10个二进制的结果，即2+(1*2+%2B+0*1)。|offset|length|style|CODE|1252978|向右移位1将sum(A_i*2%5Ei)+i=0..n+in+sum(A_(i%2B1)*2%5Ei)+i=0..n-1转换为sum(A_(i%2B1)*2%5Ei)i=0..n-1这就是为什么如果您的数字是偶数(即A_0+=+0)，它会被除以2。(对于自定义的LateX语法，很抱歉...：)|1252979|entityMap^0|0|0|0|0|0|5|7|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@$O|12|P|13|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

11 in binary is 1011

<pre><code>11 &gt;&gt; 1 
</code></pre>

means you shift your binary representation to the right by one step.

<pre><code>1011 &gt;&gt; 1 = 101
</code></pre>

Then you have 101 in binary which is 1*1 + 0*2 + 1*4 = 5. 
If you had done <code>11 &gt;&gt; 2</code> you would have as a result 10 in binary i.e. 2 (1*2 + 0*1).

Shifting by 1 to the right transforms sum(A_i*2^i) [i=0..n] in sum(A_(i+1)*2^i) [i=0..n-1]
that's why if your number is even (i.e. A_0 = 0) it is divided by two. (sorry for the customised LateX syntax... :))

blocks|key|890964|text|二进制没有十进制数的概念。它返回截断(int)值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|890965|11+=+1011，以二进制表示。向右移动，你会得到101，这是十进制的5。|890966|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Binary has no concept of decimal numbers. It's returning the truncated (int) value.

11 = 1011 in binary. Shift to the right and you have 101, which is 5 in decimal.

blocks|key|1253004|text|位移位与除法或乘以2%5En相同。在整数算术中，结果会四舍五入为整数。在浮点运算中，位移位是不允许的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1253005|在内部，位移位，嗯，移位，而舍入仅仅意味着从边缘掉下来的位被移除(并不是说它实际上会计算精确的值，然后对其进行舍入)。出现在相对边缘的新比特对于右侧和正值始终为零。对于负值，在左边附加一位，这样值就保持为负(参见two's+complement的工作原理)，并且我使用的算术定义仍然有效。|offset|length|1253006|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Two's_complement^0|0|2Y|G|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

Bit shifting is the same as division or multiplication by 2^n. In integer arithmetics the result gets rounded towards zero to an integer. In floating-point arithmetics bit shifting is not permitted.

Internally bit shifting, well, shifts bits, and the rounding simply means bits that fall off an edge simply getting removed (not that it would actually calculate the precise value and then round it). The new bits that appear on the opposite edge are always zeroes for the right hand side and for positive values. For negative values, one bits are appended on the left hand side, so that the value stays negative (see how <a href="http://en.wikipedia.org/wiki/Two&#39;s_complement" rel="nofollow noreferrer">two's complement</a> works) and the arithmetic definition that I used still holds true.

blocks|key|145052|text|在大多数静态类型的语言中，操作的返回类型是例如"int“。这就排除了小数结果，就像整数除法一样。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|145053|(关于什么是“幕后”有更好的答案，但您不需要理解这些答案就能摸索类型系统的基础知识。)|145054|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

In most statically-typed languages, the return type of the operation is e.g. "int". This precludes a fractional result, much like integer division.

(There are better answers about what's 'under the hood', but you don't need to understand those to grok the basics of the type system.)

Hey, I'm self-learning about bitwise, and I saw somewhere in the internet that arithmetic shift (>>) by one halfs a number. I wanted to test it:

<pre><code>44 &gt;&gt; 1 returns 22, ok
22 &gt;&gt; 1 returns 11, ok
11 &gt;&gt; 1 returns 5, and not 5.5, why?
</code></pre>

Another Example:

<pre><code>255 &gt;&gt; 1 returns 127
127 &gt;&gt; 1 returns 63 and not 63.5, why?
</code></pre>

Thanks.

Why arithmetic shift halfs a number only in SOME incidents?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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嘿，我正在自学关于位的知识，我在互联网上的某个地方看到了算术移位(>>)是一个数字的一半。我想测试一下：44 >> 1 returns 22, ok22 >> 1 returns 11, ok11 >> 1 returns 5, and not 5.5, why?另一个例子：255 >> 1 returns 127127 >> 1 returns 63 and not 63.5, why?谢谢。

问为什么在某些情况下，算术移位只会将数字减半？
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