Thoroughly understand bit operations - shift left, shift right

I believe you can see binary operations in various languages and frameworks . Move left 、 Move right 、&、|、^ Operation, etc. . Then this post makes you thoroughly understand shift left 、 Move right .

First of all, we should distinguish the left shift 、 That's right shift , It's very simple , Distinguish from the direction pointed by the arrow .<< Move left ,>> Move right

Move left ：

To put it simply, put the current binary , Move the whole to the left N A unit of ,N Depends on your expression .

Then take an example , And draw pictures to understand .

32 << 1 = 64.32 The left one , The answer for 64. Now let's draw a picture to understand .

32 Expressed in binary

 

hold 32 Binary system （ The blue part ） Move the whole one to the left . Get the picture below .

32 The result obtained after moving one bit to the left

0100 0000 Binary to decimal 64. So the answer is 64.

Then another topic , Let's make a small change with the previous topic .

32 << 2 = 128, The answer for 64. Now let's use drawing to understand . 

32 Expressed in binary as

 

hold 32 Binary part of （ The blue part ） Move to the left 2 Units get the following figure

Binary system 1000 0000 Convert to decimal to 128.

So every left shift operation is so troublesome , We have to convert the decimal system into binary , After moving , Then convert from binary to decimal , It's too much trouble . So we need to find the rules .

The law is ： Source operands  * 2 Of N Power （N Depends on the number of bits moved ） = The result of the move .

32 << 2  = 32 * 2^2  = 128

16 << 4 = 16 * 2^4  =  256

7 << 2 = 7 * 2^2 = 28

And so on ......

Move right ：

Simply put , Put the current binary , Move the whole to the right N The unit of , Get a new binary .

Let's use an example + Draw pictures to understand .

32 >> 1 = 16.32 Moves to the right one , The answer for 16. Now let's draw a picture to understand .

hold 0010 0000 The whole blue part moves to the right , This was 6 To represent a binary , Turned into 5 To represent a binary , Another one has been moved to the right .

So get 0001 0000 Convert it to decimal 16.

12 >> 4 = 0.12 Shift four places to the right , The answer for 0. Now let's draw a picture to understand .

  hold 0000 1100 Move right 4 position , Just put the last 4 The position has been moved out , Just at the end 4 position 1100 Expressed as 12. Move right to get 0000 0000 So the final answer is 0.

Then let's find the rules

The law is ： Source operands  / 2 Of N Power （N Depends on the number of bits moved ） = The result of the move ( Only integer part )

32 >> 2  = 32 / 2^2  = 8

17 >> 2 = 17 / 2^2 = 4.25 = 4（ Because only the integer part is taken ）

512 >> 10 = 512 / 2^10 = 0.5 = 0（ Because only the integer part is taken ）

Why is bit operation efficient

Originally, I wanted to compare with intel Develop manual level to find information to demonstrate the efficiency of bit operation . But in assembly, shifting left and right is an instruction , Like addition, subtraction, multiplication and division, it is an instruction . The corresponding instruction is intel The explanation in the development manual does not mention efficiency , Just tell developers how to use these instructions （ It can also be said that the author's ability is poor , Can't find ）.

But from a macro perspective , Whether it's left or right , We all know the rules from the above , If you don't use shift left and shift right , Their calculation steps are divided into several steps , And an operation is CPU An instruction execution cycle of , Moving left and right only requires moving N It's binary , In this way, there will be several more instruction execution cycles than using shift left and shift right directly , Maybe when there are few operations , Little difference . But the advantage of using bit operation in large-scale operation appears .

Everyone should understand , Both system development and hardware development use C And compilation , Where is their high efficiency ？ A lot of advanced packaging is missing , Reduce the number of compilations , Basically for CPU Instruction set programming . And bit operations can also be analogized in this way . For the bottom of the computer, it is composed of binary , Binary computing is more inclined to low-level Computing .

summary

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