位移X *一个数字 [英] Bit shifting x * a number

问题描述

你怎么样 -10 数从这些位转移的实践问题？

据我了解 X * 32 可以写成 X＆LT;＆LT; 5 。但是你怎么让喜欢数x * 66 或 X *（ - 10）？

解决方案

常规说明

位移动是主要目的转向数字的二进制重新presentation。它的不可以乘法。

  23 = 0001 0111
23℃;＆下; 1 = 0001 0111 LT;＆LT; 1 = 0010 = 1110 46
 

  

然而，由于一个数的二进制重新presentation被改变，则
  数它重新presents也发生了变化。这是计算机的二进制系统是如何工作的。因而人们有时会利用这种行为为黑客，主要是为了加快计算时间。

让我们试着去了解它更多：

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左位移位和右位移

现在，当重新presented数量整数输入，然后通过1移一个数的二进制重新presentation到左将等同于乘法就以2：

  23 = 0001 0111
23℃;＆下; 1 = 0001 0111 LT;＆LT; 1 = 0010 = 1110 // 46左1位移，数变为一倍
 

由于有没有溢出作为给定的数据类型：

  255 = 1111 1111 //假设8位数据类型
255 LT;＆LT; 1 = 1111 1111 LT;＆LT; 1 = 1111 1110 = 254 //不乘以2，因为溢出，
 

虽然转移整数的右侧将相当于为分它由2，然后四舍五入下来

  23 = 0001 0111
23 GT;＆GT; 1 = 0001 0111 GT;＆GT; 1 = 000 1011 = 11 //右移1位移，数变为减半，舍去
 

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有些使用，并链接到乘除法

由于位换档操作通常比乘法成本更低，加快速度，你会在一些项目看，人用左位移位操作（如更换乘法）时，他们的意思<$乘它C $ C> 2 （即2，4，8，16等）功率的整数：

  int类型的= 23;
...
A = A＆所述;＆下; 2; // = 102;乘以4，当量为a = A * 4，但更快的操作
 

或者使用权的比特移位操作（作为替代除法和向下舍入的），以将其与划分的功率的整数2 （也就是2,4 ，8，16，等）

  int类型的= 23;
...
A = A＆GT;＆GT; 2; // = 5;除以4，并四舍五入，相当于整数除法A = A / 4，但速度更快
 

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结束语

请注意，只有当你用数字与2电源运行，所有的乘法和上述部门可以通过左位移位或右侧位移来代替。

在你的榜样，66和-10不是整数，其中2的功率，所以你不能黑客二进制移操作乘法/除法。

在一般情况下，使用位移位，如果你意味着位转移，如位移有许多其他用途的不仅仅是黑客的乘法/除法与2的幂的整数如果你想乘操作或鸿沟，很乐意与只用乘法（ * ）或除法（ / ）运算符。

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一些附加备注：

话虽这么说，我只是想补充有关的位移进一步解释一些更多的东西（它不会伤害）：

• 签署整数键入可以保持正或负号


• 有逻辑比特移位和具有负号处理时算术位移之间的差。一会给 0 在清空空间后移，而另一会给 1


• 因此，它可能是最好的注意，位移位主要用于无符号类型，比如像点点转移创造位掩码。也就是说，无符号建议用于避免符号扩展惊喜，当你处理负数（右）位移位。

How do you get number like -10 from these bit shifting practice problems?

From what I understand X*32 can be written as x<<5. But how are you to get numbers like x*66, or X*(-10)?

解决方案

General Explanation

Bit shifting is primarily aimed to shift the binary representation of a number. It is not for multiplication.

23 = 0001 0111
23 << 1 = 0001 0111 << 1 = 0010 1110 = 46

However, as the binary representation of a number is changed, the number it represents is also changed. This is just how computer binary system works. And thus people sometimes exploit this behavior as a "hack", mostly to speed up the computation time.

Let's try to understand it more:

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Left bit-shift and right bit-shift

Now, when the number represented is of integer type, then shifting the binary representation of a number by 1 to the left will be equivalent to multiplying it by 2:

23 = 0001 0111
23 << 1 = 0001 0111 << 1 = 0010 1110 = 46 //left bit-shift by 1, number becomes doubled

Given that there is no overflow for the given data type:

255 = 1111 1111 //assuming 8-bit data type
255 << 1 = 1111 1111 << 1 = 1111 1110 = 254 //not multiplied by 2, because of overflow

While shifting integer number to the right will be equivalent as dividing it by 2 and then rounding it down:

23 = 0001 0111
23 >> 1 = 0001 0111 >> 1 = 000 1011 = 11 //right bit-shift by 1, number becomes halved, rounded down

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Some use and link to multiplication and division

Since bit-shifting operation is typically less costly than multiplication, to speed things up, you will see in some program, people use left bit-shift operation (as a replacement of multiplication) when they mean to multiply it by an integer number of power of 2 (that is 2, 4, 8, 16, etc):

int a = 23;
...
a = a << 2; //=102; multiply by 4, equivalent to a = a * 4, but faster operation

Or use right bit-shift operation (as a replacement of division and rounding down) to divide it with an integer number of power of 2 (that is 2, 4, 8, 16, etc)

int a = 23;
...
a = a >> 2; //=5; divide by 4 and rounding down, equivalent to integer division a = a / 4, but faster

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Concluding remarks

Note that only if you operate with number with power of 2, all the multiplications and divisions above can be replaced by left bit-shift or right bit-shift.

In your example, 66 and -10 are not integer number which of power of 2, thus you cannot "hack" the multiplication/division with binary-shifting operation.

In general, use bit-shift operation if you mean for bit-shifting, as bit-shifting has many other uses than just "hacking" for multiplication/division with integer number of power of 2. If you want to multiply or divide, be happy with just using multiplication (*) or division (/) operator.

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Some additional remarks:

That being said, I would just like to add some more things regarding the bit-shift for further explanation (it won't do harm):

• signed integer type can hold positive or negative number
• there is a difference between logical bit-shift and arithmetic bit-shift when dealing with negative number. One will give 0 in the emptied-space after shift while the other will give 1
• Hence, it probably best to note that the bit-shift is mainly used for unsigned type, such like for creating bit-masks by bit shifting. That is, unsigned is recommended to be used to avoid sign-extension surprises when you deal with negative number (right) bit-shift.

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