Bit shifting x * a number

问题

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)?

回答1:

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.

来源：https://stackoverflow.com/questions/35762313/bit-shifting-x-a-number